a ball is thrown straight up from the ground with speed v0 . 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.

When a ball is thrown straight up from the ground with an initial speed v0v_0v0​, it moves upward, decelerating due to gravity until it reaches its maximum height where its velocity is zero, then accelerates downward under gravity

. At the same instant, a second ball is dropped from rest from a height hhh directly above the first ball's launch point. Assuming no air resistance, both balls experience the same gravitational acceleration g=9.8 m/s2g=9.8,m/s^2g=9.8m/s2 downward throughout their motion

. The position of the first ball at time ttt after being thrown upward is:

y1(t)=v0t−12gt2y_1(t)=v_0t-\frac{1}{2}gt^2y1​(t)=v0​t−21​gt2

The second ball, dropped from height hhh with zero initial velocity, falls downward with position:

y2(t)=h−12gt2y_2(t)=h-\frac{1}{2}gt^2y2​(t)=h−21​gt2

The two balls collide when their positions are equal, i.e., y1(t)=y2(t)y_1(t)=y_2(t)y1​(t)=y2​(t). Setting the equations equal:

v0t−12gt2=h−12gt2v_0t-\frac{1}{2}gt^2=h-\frac{1}{2}gt^2v0​t−21​gt2=h−21​gt2

This simplifies to:

v0t=hv_0t=hv0​t=h

Therefore, the time ttt when they collide is:

t=hv0t=\frac{h}{v_0}t=v0​h​

At this time, the collision height is:

y=v0t−12gt2=h−12g(hv0)2y=v_0t-\frac{1}{2}gt^2=h-\frac{1}{2}g\left(\frac{h}{v_0}\right)^2y=v0​t−21​gt2=h−21​g(v0​h​)2

This shows the collision occurs below the starting height hhh, but above the ground. In summary:

• The first ball moves up and then down under gravity with initial speed v0v_0v0​.
• The second ball falls from rest from height hhh.
• They collide at time t=hv0t=\frac{h}{v_0}t=v0​h​.
• The collision height is less than hhh but greater than zero.

This analysis assumes no air resistance and uniform gravitational acceleration