#Linear Motion
101 messages · Page 1 of 1 (latest)
Try using v² - u² = 2as during both journeys.
I write d instead of s, its the distance
4 - 0.25 = 2 x a x 2
3.25 = 2 x a x 2
a = 0.8125 m/s^2
Hmmm? No no. You squared distances
It has been made impossible to find velocity. @everyone
Velocity depends on distance covered and time.
You can find tho.
OK
Who has got an answer apart from @woven flume @everyone ?
Why don't you try applying those suvat stuff
What suvat stuff?
Why are you pinging everyone
@everyone, who has found the answer?
There is like 20k people
I only need an immediate answer to my question that's all.
Equations of motion
Please help, I'm an amateur
Accelerates during takeoff to get some velocity. Which would be initial velocity for journey without external acceleration other than g,
v² - u² = 2ad
0 - u² = 2(-g)(1.5)
u = sqrt(3g)
So, velocity gained through accelerated takeoff is sqrt(3g)
Again, v² - u² = 2ad
3g - 0 = 2a(0.5)
But deceleration occurs during upward motion.
So she accelerated at 40 m/s^2
Yeah that's why -g
30?
No 40
I don't see how it'd be 40
Yeah I know.
But It is it.
We are talking about deceleration.
You probably went wrong somewhere
Where?
I wouldn't know unless you show your working on how you got 40
So that is where I went wrong?
So when substituted with 0.5 m as the distance, the acceleration is 40 m/s.
Why is initial velocity 2sqrt10
Because the deceleration is 10 m/s.
Initial velocity as in at what point of journey.
There is no journey.
As in, the person wasn't moving at all?
In the question, Valerie moved upwards to spike the ball.
Yeah, so corresponding to what height the velocity is 2sqrt10?
Corresponding to 2 metres hight.
2 metres is the maximum height. Where velocity is 0.
I mean the constant velocity.
Hmmm idk what you mean by constant velocity. Cuz velocity is changing constantly.
Constant velocity is when the rate of change of displacement with time remains equal.
This question is a challenging one so it would need much analysis.
Which is not a case in this question so that's irrelevant
But I found the answer to be 40 m/s^2
Tbh it's not. I'd actually advise you to practice basic questions thoroughly so you have good conceptual understanding
You haven't given appropriate explanation to it
Guess you did v = sqrt(2×10×10) to get v = 2sqrt10. But that would be the case when it falls freely or thrown up through 2 meters. Question says that 0.5 m of that 2 is accelerated upwards.
@summer epoch you probably set as acceleration a-g
Where a the acceleration due to the jump force
The total acceleration is 30 m/s²
The part coming from the jump force is 40m/s²
The question asks for the total so the answer is 30
@mild notch and @tall owl, OK.
Nothing was thrown up.
Let's say she was thrown up. You lose nothing by saying that.
Assumption is not accurate.
She rose to spike the ball with the hight of 2 m and accelerated 0.5 m.
Bro are you concerned about the literature aspect
Yeah.
I'd call it immature
Let us not assume things for we would be wrong.
It was an analogy you don't lose any information required for the answer
It's here
How can you initial and final velocity be the same?
There is no understanding in what you have done there.
But at least you did your best. Thank you for that.
Velocity when h=0.5 m is sqrt(30) m/s
OK.
I missed a m/s^2 square
You and Mark as read are giving me one answer.
Yeah thats what the question asks
OK.
Are you aware you are you used 1.5 m to find the velocity?
Look @tall owl
√(2)(g)(h)=√(2)(1.5)(10)= √30 m/s.
I think you are supposed to use the total distance 2 m to find the velocity.
Yeah h2-h1=1.5
But shouldn't you use the total distance?
No the conservation of energy is done from point 1 to point 2, h1=0.5 m , h2=2 m
OK