physics

[lee92 2]

i know this probly won't aid my case, but i've spent about 4 hrs puzzleing over the answer to this physics question, and frankly i'm losing my mind over it. can anyone please help me!?

'Sitting in a second story apartment, a physicist notices a ball moving straight upward just outside her window. The ball is visible for 0.292s as it moves a distance of 1.04m from the bottom to the top of the window. What is the greatest height of the ball above the top of the window? How long does it take before the ball reappears?'

[Cavs019 710]

Do they give you anything else? Velocity, window height in relation to the ground?

[asdfa 0]

okay, so you know the average velocity of the ball as it moves from the bottom of the window to the top as being 1.04m/0.292s or around 3.56 m/s.

that is the average velocity, it's v1+v2 = v average.

since accelerate is constant at -9.8m/s^2, the change in velocity is 0.292s x -9.8 m/s^2

so the acceleration on the ball results a change in velocity of 2.8616 m/s.

the average velocity (3.56 m/s) = (velocity 1, x + velocity 2, x - 2.8616)/2

(3.56 m/s + 1.4308m/s) = x

4.9908m/s = x

velocty 2 is then 4.9908m/s - 2.8616 m/s = 2.1292 m/s

so now you have to find out how far the ball travels past the height of the window, at which point the ball's velocity will be 0.

therefore average velocity will be (2.1292m/s + 0 m/s)/2 = 1.0646m/s.

then find out how long it takes the ball to deccelerate from 2.1292m/s to 0

2.1292m/s / 9.8 m/s^2 = 0.21726s

SO, the distance travelled by the ball past the window is 2.1292 m/s x 0.21726s = 0.23129m.

Since the acceleration of gravity is constant, the ball will take the same time coming down as it did going up, so the time until the ball reappears is

0.21726s x 2 = 0.43452s

uh, at least thats what i think it is, check the math and stuff. it's been a while since i've done stuff like this

hope i helped

[Chadd 916]

I'm not great at the formula part but I believe this is the methodology:

Determine the speed of the object

Determine how far it will go before coming to a stop (gravity working against it)

Determine how long it will take to come to a stop

Determine how long for that object to fall the distance back down to the window

*****

Beat me to it, damn customers

[lee92 2]

okay, so you know the average velocity of the ball as it moves from the bottom of the window to the top as being 1.04m/0.292s or around 3.56 m/s.

that is the average velocity, it's v1+v2 = v average.

since accelerate is constant at -9.8m/s^2, the change in velocity is 0.292s x -9.8 m/s^2

so the acceleration on the ball results a change in velocity of 2.8616 m/s.

the average velocity (3.56 m/s) = (velocity 1, x + velocity 2, x - 2.8616)/2

(3.56 m/s + 1.4308m/s) = x

4.9908m/s = x

velocty 2 is then 4.9908m/s - 2.8616 m/s = 2.1292 m/s

so now you have to find out how far the ball travels past the height of the window, at which point the ball's velocity will be 0.

therefore average velocity will be (2.1292m/s + 0 m/s)/2 = 1.0646m/s.

then find out how long it takes the ball to deccelerate from 2.1292m/s to 0

2.1292m/s / 9.8 m/s^2 = 0.21726s

SO, the distance travelled by the ball past the window is 2.1292 m/s x 0.21726s = 0.23129m.

Since the acceleration of gravity is constant, the ball will take the same time coming down as it did going up, so the time until the ball reappears is

0.21726s x 2 = 0.43452s

uh, at least thats what i think it is, check the math and stuff. it's been a while since i've done stuff like this

hope i helped

thank you! i've tried a dozen answers, but have yet to get that one. i'll work it out n put it in n see if she's right! thank you again!