Posted by **Anonymous** on Friday, May 28, 2010 at 1:47pm.

A ball is thrown straight up from the ground with speed v_0. At the same instant, a second ball is dropped from rest from a height H, directly above the point where the first ball was thrown upward. There is no air resistance.

Find the value of H in terms of v_0 and g so that at the instant when the balls collide, the first ball is at the highest point of its motion.

- Physics -
**bobpursley**, Friday, May 28, 2010 at 3:14pm
The highest point is when vertical velocity is zero.

Vf^2=Vi^2+2gh

h= vi^2/2g= Vi^2/19.8

Now, you need for the top ball to fall H-h in the same time.

time= distance/avgvelocityfor the bottom ball

= Vi^2/2g*Vi/2= Vi/g

now to the top:

H-h=1/2 g t^2= 1/2 g (Vi^2/g^2)=1/2 Vi/g

Now solve for H.

check all that.

## Answer This Question

## Related Questions

- Physics - A ball is thrown straight up from the ground with speed . At the same ...
- physics - The average velocity for a trip has a positive value. Is it possible ...
- physics - A ball is thrown straight upward and returns to the thrower's hand ...
- Physics - The Colliding Balls: A ball is released from rest at a height h above ...
- physics - An object is thrown straight up. At the same instant that the object ...
- Physics - A rock is dropped from a height of 100 meters above the ground. One ...
- Physics - A rock is dropped from a height of 100 meters above the ground. One ...
- physics - A ball is thrown upward from the ground with an initial speed of 25 m/...
- Physics - A ball is thrown upward from the ground with an initial speed of 25 m/...
- physics - A ball is thrown upward from the ground with an initial speed of 26.2 ...

More Related Questions