Does Velocity Remain Constant in Vertical Motion?

[revernance]

A need some help understanding the concept:

A person on a trampoline bounces straight upward with an initial speed of 4.0 m/s.

What is the person's speed when she returns to her initial height?

The answer is 4.0 m/s.

Can you guys explain why though?
Thanks so much!

 

[arildno]

Conservation of mechanical energy.
At the top, all kinetic energy had been converted into potential energy; returning, that potential energy re-converted into kinetic energy.

 

[Doc Al]

What do you think is going on?

Hint: It's got nothing to do with trampolines. If you threw a rock straight up into the air with a speed of 4 m/s, what speed will it have when it falls back into your hand? (Ignoring air resistance, of course.)

 

[jwalker1196]

Maybe what's confusing you is you're thinking of friction. The equation which produces that result ignores friction (which includes wind resistance, etc). So in the real world, no, you don't land with quite as much speed.

Also don't forget the difference between velocity and speed. Initial speed = final speed, sure. But initial velocity, +4.0 m/s in the y-axis (for trampoline example) yields -4.0m/s in the y-axis for final velocity.

 

[saiaspire]

gravity is a conservative force/./

when u throw the body and it reaches the max height, its velocity is 0( so KE is 0), whereas it has PE( which is = to mgh)

when it reaches back to the point of projection, its KE will be equal to the PE at highest point( since gravity is conservative force). therefore , the velocity will be equal to the velocity with which u throw...

 

[saiaspire]

or using Newtons laws..

initial velocity = u

at highest point v=0;

v=u+at;
a=g=10 m/s^2
u=10t;

now from highest point.

u=0;
v=0-10t(- ve for direction);
v= -10t;