mechanics

[red crown]

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[tulip burrow]

You have a point O from where a ball is fired. The height above O is satisfied by h = $\frac{1}{20}(18x - 3x^2)$
\newline\newline
a) When it's again at same level as O, h will be zero. Hence solve for x.
\newline\newline
b) i) just substitute x=2 and solve for h.
\newline\newline
ii) Use distance formula (coordinate geometry) taking O to be (0,0)

Did I make it clear?

[real sphinx]

Dash

[fallen gust]

yes for a) and b) i) but for ii) how would you find the x displacement?

would it be the x value from a)?

\newline\newline
a) x=-6
\newline\newline
b) i) h=1.5m
\newline\newline
ii) $(1.5)^2 + (6)^2= distance^2$

[real sphinx]

Goose
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[fallen gust]

@tulip burrow

[tulip burrow]

its given right un the question, x = 2.

and your answer for a and b i) are wrong too. Check again.

a) Why would it be negative 6? Check the math.
b) i) You just have to substitute x = 2 and solve for x.
ii) Given in the question x=2 and you found y from b i). Use distance formula as I've already said.

[fallen gust]

$0=1/20(18x-3x^2)$
\newline\newline
$0=18x-3x^2$
\newline\newline
$0=18-3x$
\newline\newline
$-18=3x$
\newline\newline
$x=-6$

[real sphinx]

Goose

[fallen gust]

for b) i) h=1.2m
\newline\newline
ii) $(square root of (1.2)^2 +(36)^2)= 6.1m$

[real sphinx]

Goose

[tender stratus]

you can use \verb+\sqrt+ for square roots, like $\sqrt{5}$

[real sphinx]

vin100

you can use \verb+\sqrt+ for square roots, like $\sqrt{5}$

[fallen gust]

thanks for the advice. is the answers i got right though?

[analog forum]

@fallen gust has given 1 rep to @tender stratus

[tender stratus]

generic skill for proofreading: substitute your answer back to the given equation to verify that it's a solution

you can use ,calc on #bot-cmd-latex to self-check your answer

,calc x = -6; 18 * x - 3 * x^2

[real sphinx]

Result:

[-216]

[tender stratus]

so your answer is wrong

use the bisection strategy to spot out where your solution goes wrong

1. the top line is good
2. the bottom line is bad
3. find the middle line. verify whether your wrong answer solves the equation.
4. if yes, then this line is bad, focus on the first half of your solution.
5. otherwise, this line maybe good, focus on the later half of your solution.
6. repeat this search process until you have only one line left

take your solution as an example, your 3rd line maybe good

so i focus on

the 4th line

[jagged flume]

Erm whil vin is offline i guess i can help if you guys need help

@fallen gust

[fallen gust]

x= 6

how would i do part c)

[fallen gust]

ty for the offer

[analog forum]

@fallen gust has given 1 rep to @jagged flume

[tender stratus]

you mean part (b)(ii)?

[fallen gust]

yes thats the one

[fallen gust]

hers what im thinking:
\newline\newline
-We are given x=2
\newline\newline
-To find the 'y' value:
\newline\newline
Put 'h' equal to 'y', making y=1.2m
\newline\newline
Then use the distance formula to find 'd':
\newline\newline
$\sqrt{1.2^2 + 2^2} = d$

[real sphinx]

Goose

[tender stratus]

,calc x = 2; (18 * x - 3 * x^2) / 20

[real sphinx]

Result:

[1.2]

[tender stratus]

calculations LGTM.
when you introduce new variables (not defined in the problem statement), you need to explain what it means (in words)
write less, so you have more time to proofread, and your grader can give you marks quicker.

[tender stratus]

you don't have to write 'y' or 'd'

just say

,,\textrm{required distance} = \sqrt{x^2 + h^2} = \dots

[real sphinx]

vin100

[tender stratus]

x and h are defined in the problem statemetn

statement*

[fallen gust]

i see

[fallen gust]

thanks for the help

[analog forum]

@fallen gust has given 1 rep to @tender stratus

[fallen gust]

+close