## distance unknown, have acceleration time and another distance

1. The problem statement, all variables and given/known data

A rock is thrown up past a window. The rock requires 0.20 s to completely pass by the 1.4 m window calculate the height above the window the rock would travel.

2. Relevant equations

D=vf(t) -1/2at^2

3. The attempt at a solution

I have tried using the acceleration as 9.81 m/s and 7 m/s and the vf as i got that from 1.4/0.20. It does not give me the right answer though. any help would be great!
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 Recognitions: Gold Member What you are given is time for completing 1.4m is 0.2s, can you find the initial velocity of stone using this data?
 So i found that the initial velocity is 6.019 m/s now how does that lead me to how far above the window the rock will travel?

Recognitions:
Gold Member

## distance unknown, have acceleration time and another distance

I don't get my answer to be 6.019m/s.

Show me the steps you followed.

 Quote by Pranav-Arora I don't get my answer to be 6.019m/s. Show me the steps you followed.
1.4m=vi(.20)+1/2(9.81)(0.20)^2
1.4=vi(0.20)+.1962
1.2038=vi(0.20)
vi=6.019m/s

Recognitions:
Gold Member
 Quote by vicsic 1.4m=vi(.20)+1/2(9.81)(0.20)^2 1.4=vi(0.20)+.1962 1.2038=vi(0.20) vi=6.019m/s
Ah...there you go wrong.
The direction of acceleration which is g in our case is opposite to the initial velocity.
Initial velocity is in upward direction and acceleration is in downward direction. So one has to be negative. That means you didn't take care of signs. If you choose upward as positive then downward direction has to be negative.

Try again & post the steps here.

 Quote by Pranav-Arora Ah...there you go wrong. The direction of acceleration which is g in our case is opposite to the initial velocity. Initial velocity is in upward direction and acceleration is in downward direction. So one has to be negative. That means you didn't take care of signs. If you choose upward as positive then downward direction has to be negative. Try again & post the steps here.

1.4m=vi(0.20)+1/2(-9.81)(0.20)^2
1.4m=vi(0.20)-0.1962
1.5962=vi0.20
vi=7.981 m/s

then plugging it back in

d=vi(t)+1/2at^2
d=7.981(0.20)+1/2(9.81)(0.20)^2
d=1.7924
d=1.8m

thanks for the help :)

Recognitions:
Gold Member
 Quote by vicsic Got the answer! 1.4m=vi(0.20)+1/2(-9.81)(0.20)^2 1.4m=vi(0.20)-0.1962 1.5962=vi0.20 vi=7.981 m/s then plugging it back in d=vi(t)+1/2at^2 d=7.981(0.20)+1/2(9.81)(0.20)^2 d=1.7924 d=1.8m thanks for the help :)
You're still incorrect. :)

When you substituted the value of vi back into the same equation, you did a sign mistake again.

Moreover, you aren't asked the initial velocity in the question.
You are asked how high it would go above the window.
For that you need to find the maximum height the stone can achieve.
(Hint: Velocity at maximum height is 0m/s).

Try again and post the steps here. :)

 Tags acceleration, distance, kinematics, velocity