#question about trajectory of object from a height

351 messages · Page 1 of 1 (latest)

rugged hare
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A stone is thrown with a velocity of 19.6 metre per second at an angle of 30 above horizontal from the top of building 14.7 metre height find the time after which the stone strike the ground and the distance of the landing point of the stone from the building and the velocity with which the stone its diagram and the maximum height attained by the stone above the ground take G is equals to 9.8 metre per second square

median zodiacBOT
valid mango
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I assume we're treating this as 2D kinematics with free fall? (no air resistance)

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If so, what have you done already?

rugged hare
valid mango
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okay, let's start with the horizontal distance traveled

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assuming that your Time of Flight is correct

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what do we know about horizontal movement in this type of problem?

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(i.e. the acceleration, delta x, time, initial velocity, end velocity)?

rugged hare
valid mango
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yes (is u = v_0?)

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just getting notation down properly

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but assume so, then yes

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how can we use that information to get our initial horizontal and vertical speeds?

rugged hare
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ive tried using the formula
R=u^2sin2theta/g
is that applicable?

valid mango
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I'm, not fully sure I understand what that formula is

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can we use use a standard trig triangle here?

rugged hare
rugged hare
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cant get it properly in text 💀

valid mango
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oh

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I've... never seen this before

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but let's see if it works

rugged hare
rugged hare
valid mango
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oh

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here's an issue

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it doesn't account for the fact that we're on top of a building

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like anywhere in the equation

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so it's a bit off (if I've done my maths right)

rugged hare
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how do we proceed

valid mango
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sorry for it being sideways > - <

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but how might we find the two other sides for this triangle (you probably know it hehe, I just need to check)

rugged hare
rugged hare
valid mango
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yeaah

rugged hare
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rightt

valid mango
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so using trig, what is our intial horizontal velocity?

rugged hare
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1/2 / 19.6

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shi wait

valid mango
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is that veritcal?

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yeah you good lmao

rugged hare
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one sec

valid mango
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also, as a fun note, are you learning AP physics 1 on your own time?

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because the AP season just ended 2 days ago (scores came out)

rugged hare
rugged hare
valid mango
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correct!

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nice

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also so now let's go back to the problem

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if we know that there is NO air resistance

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does the acceleration change at all for the rock (ball?)?

rugged hare
valid mango
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nice

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so if a is constant

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what is a?

rugged hare
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9.8

valid mango
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horizontally?

rugged hare
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g innit

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ohhh

valid mango
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do we have horizontal gravity?

rugged hare
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NO

valid mango
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WE MOVING FAM

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yeah there's none

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so what do we know now?

rugged hare
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acceleration horizontal is 0

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yeyey

valid mango
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$u=19.6\cdot\sqrt{3}/2$

zenith totemBOT
valid mango
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a = 0

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our time in air is 3 seconds

rugged hare
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sub values in equation 1 of motion?

valid mango
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and what can we use with these three things to find our horizontal distance traveled?

rugged hare
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OHNO

valid mango
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uhhhhh

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which one is that

rugged hare
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s=ut+1/2at^2?

valid mango
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nice

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yes

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plug it in and see what you get

rugged hare
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alrrr

valid mango
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should be around 50.92 metres

rugged hare
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alr im getting 50.9 (approx) taking root3 as 1.73

valid mango
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yeahhhhhhhhhh

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alright that's 2 out of like 5 done?

rugged hare
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yessirr

valid mango
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okay 2/4

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so which one do you want to do next?

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either works

rugged hare
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v at which it lands?

valid mango
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aight that works

rugged hare
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alr

valid mango
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so

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we know that at the start, the rock is going UP

rugged hare
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yeas

valid mango
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actually

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max height might be easier

rugged hare
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alr lets go with that

valid mango
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since from my 3 brain cells, that's how I would get max velocity

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so what characteristic of max height might help?

rugged hare
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?

valid mango
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is all velocity = 0?

rugged hare
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final velocity

valid mango
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or just horizontal?

rugged hare
valid mango
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or just vertical?

rugged hare
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vertical

valid mango
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yes

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vertical v = 0

rugged hare
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yeye

valid mango
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so let's remove the horizontal aspect of the question out

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and get the vertical done

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in our initial speeds

rugged hare
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alr

valid mango
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what is our intial vertical speed

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using trig

rugged hare
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root3 19.6/2

valid mango
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... is that the same as the horizontal one?

rugged hare
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im not sure 💀

valid mango
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we're not in a 45 45 90 triangle, so I would assume vertical and horizontal velocity is different

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back to the picture!

rugged hare
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alrr

valid mango
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so in my crudely sketched diagram

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the "how find?" portion is our vertical speed

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LMAO

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so would that be sine or cosine?

rugged hare
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sine

valid mango
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YES

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and what is sine 30 deg?

rugged hare
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1/2

valid mango
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nice

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so what is vertical

rugged hare
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1/2=x/19.6

valid mango
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what

rugged hare
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9.8

valid mango
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yes

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okay

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we good

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so

rugged hare
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yessor

valid mango
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in our kinematics equation

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we know accelertaion is what?

rugged hare
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9.8

valid mango
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yes

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so vertical velocity = 9.8

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acceleration = 9.8

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and end velocity = 0

rugged hare
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right

valid mango
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but is this correct?

rugged hare
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i believe so?

valid mango
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(velocity and acceleration are vectors)

rugged hare
valid mango
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in my statements, I said that we started going upward

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but gravity points downwards

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so what needs to change in the variables?

rugged hare
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yes

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-9.8/

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?

valid mango
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YES

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acceleration is negative here

rugged hare
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RIGHT

valid mango
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so using our equation

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which one helps us find elapsed distance?

rugged hare
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s=ut+1/2at^2

valid mango
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correct again

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yeah we use that a lot 😭

rugged hare
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or could we use v^2=u^2+2as

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cuz then v=0?

valid mango
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oh

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yeah that's right

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woopsies

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im dumb

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ohhhhhh

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yeah

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so we have initial velocity = 9.8

rugged hare
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yey7e

valid mango
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end velocity = 0

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acceleration = idrk

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g

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wait did i do this right

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one sec

rugged hare
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-9.8 innit

valid mango
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I haven't done physics since the AP exam sob

rugged hare
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rightt

valid mango
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okay

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it's

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neither of those equations i think

rugged hare
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😲

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go on

valid mango
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i think it's 5

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where we have

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acceleration

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displacement

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inital velocity

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end velocity

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looks like

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$2a\cdot\delta x=v_2^2-v_1^2$

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$2a\cdot\Delta x=v_2^2-v_1^2$

zenith totemBOT
valid mango
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there we go

rugged hare
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what zehell

valid mango
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delta x is your s

rugged hare
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im sorry but im afraid ive never seen this equation in my life

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💀

valid mango
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💀

rugged hare
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im listening tho go on

valid mango
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huh

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well

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let's see what you have

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because I don't wnat to throw a random thing at you

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and call it a day

rugged hare
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alr alr

valid mango
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what equations do you have?

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that include

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initial velocity

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end velocity

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acceleration

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and displacement

rugged hare
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ye v=u +at
s=ut +1/2at^2
v^2=u^2+2as

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i think the 3rd matches

valid mango
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okay

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let's work with that then

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try plugging in your numbers

rugged hare
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im getting 4.9?

valid mango
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i'm also getting 4.9

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so we're running with that hahah

rugged hare
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alrighhttt

valid mango
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LAST ONE

rugged hare
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the final one alas

valid mango
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so

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we have our position in space right?

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if we started 14.7 meters up

rugged hare
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yes

valid mango
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and we've gone 4.9 meters up to the new point

rugged hare
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yes

valid mango
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so 19.6 meters high is our new position

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and we know that the end horizontal velocity is what?

rugged hare
valid mango
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let's review then, we chillin

rugged hare
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alrightt

valid mango
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so initial horizontal is what?

rugged hare
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19.6?

valid mango
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no that's combined initial speeds

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this is after we used cosine

rugged hare
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OH

valid mango
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yee

rugged hare
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root3 19.6/2

valid mango
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yes

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and our horizontal acceleration?

rugged hare
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0

valid mango
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so is it the same?

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before and after?

rugged hare
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yes?

valid mango
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YES

rugged hare
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alrightt

valid mango
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our horizontal velocity does not change

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so now we just need to get vertical velocity

rugged hare
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right

valid mango
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we are 19.6 meters in the air

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gravity = 9.8 = acceleration

rugged hare
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yes

valid mango
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end velocity is what?

rugged hare
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19.6 root3 /2

valid mango
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wait

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that's what we're solving for

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oopsies

rugged hare
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right

valid mango
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what is our INITIAL vertical velocity?

valid mango
rugged hare
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initial vertical is 19.6?

valid mango
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that's our height

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we ended the previous question with this

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what characteristic of vertical velocity gave us the max height?

rugged hare
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vfinal=0

valid mango
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yes

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so our new velocity

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at that same height

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is what?

rugged hare
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OHH

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0

valid mango
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YEEAH

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so

rugged hare
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RIGHT

valid mango
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initital velocity is 0

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acceelration is 9.8

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displacement is 19.6

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solving for end velocity

rugged hare
valid mango
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so since we're planning for the rock to go down

rugged hare
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positive

valid mango
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and our gravity is already going down

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we can use poisitive

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yeah

rugged hare
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rightt

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384.1 6?

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wtf

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OH

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nah bro im tripping
i got 19.6 again?

valid mango
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yteah

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i dont like this problem anymore

rugged hare
valid mango
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there's so many duplicates 😭

rugged hare
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ikrr

valid mango
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yeah but i get 19.6

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so now

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final step

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our end vertical is 19.6 m/s

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our end horizontal is 16.97 m/s (the sqrt 3 stuff)

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what can we use here

rugged hare
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im not sure
can you brief

valid mango
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okay

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diagram time!

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one sec

rugged hare
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💯

rugged hare
valid mango
rugged hare
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right

valid mango
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do u see?

rugged hare
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give me a sec

valid mango
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okay

rugged hare
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pythagoreas theoreom?

valid mango
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yep

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and HTAT'S IT

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yaaaaay

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you'll get used to these style of problems after enough practice

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it's tedious like that

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Time in air: 3 seconds

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Horizontal Displacement: 50.92 metres

rugged hare
rugged hare
valid mango
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Max Height: 4.9 meters above 14.7 = 19.6 m

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and your last part

rugged hare
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one sec

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25 approx

valid mango
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yeah that works

rugged hare
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26

valid mango
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I got 25.92

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or 93

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good job!

rugged hare
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ALRIGHT

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thank you very much mate

valid mango
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if you need anymore help, feel free to put in another question

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but bye bye!

rugged hare
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.close