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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.

They collide at time t=hv0t=\dfrac{h}{v_0}t=v0​h​, at a height 12(h+v02g)\dfrac{1}{2}\left(h+\dfrac{v_0^2}{g}\right)21​(h+gv02​​) above the ground, provided the collision happens while both are in flight.

Set up the motion

Take upward as positive and the ground as y=0y=0y=0.

  • Ball 1 (thrown up from ground):
    Initial position y1(0)=0y_{1}(0)=0y1​(0)=0, initial velocity v0v_0v0​.
    Its position is

y1(t)=v0t−12gt2.y_1(t)=v_0t-\tfrac{1}{2}gt^2.y1​(t)=v0​t−21​gt2.

  • Ball 2 (dropped from height hhh):
    Initial position y2(0)=hy_{2}(0)=hy2​(0)=h, initial velocity 000.
    Its position is

y2(t)=h−12gt2.y_2(t)=h-\tfrac{1}{2}gt^2.y2​(t)=h−21​gt2.

Time of collision

At collision, they are at the same height, so y1(t)=y2(t)y_1(t)=y_2(t)y1​(t)=y2​(t).

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

The −12gt2-\tfrac{1}{2}gt^2−21​gt2 terms cancel, leaving:

v0t=h⇒t=hv0.v_0t=h\quad \Rightarrow \quad t=\frac{h}{v_0}.v0​t=h⇒t=v0​h​.

So the collision time is

tcollision=hv0.t_{\text{collision}}=\frac{h}{v_0}.tcollision​=v0​h​.

Height of collision

Substitute t=hv0t=\dfrac{h}{v_0}t=v0​h​ into either position equation, say y2(t)y_2(t)y2​(t):

ycollision=h−12g(hv0)2=h−gh22v02.y_{\text{collision}} =h-\tfrac{1}{2}g\left(\frac{h}{v_0}\right)^2 =h-\frac{gh^2}{2v_0^2}.ycollision​=h−21​g(v0​h​)2=h−2v02​gh2​.

Or into y1(t)y_1(t)y1​(t):

ycollision=v0(hv0)−12g(hv0)2=h−gh22v02,y_{\text{collision}} =v_0\left(\frac{h}{v_0}\right)-\tfrac{1}{2}g\left(\frac{h}{v_0}\right)^2 =h-\frac{gh^2}{2v_0^2},ycollision​=v0​(v0​h​)−21​g(v0​h​)2=h−2v02​gh2​,

which matches, as it must.

This can be rewritten in a symmetric form if one relates hhh and v0v_0v0​ for particular problems, but the key results for this setup are:

  • Time when they collide: t=hv0t=\dfrac{h}{v_0}t=v0​h​.
  • Height where they collide: y=h−gh22v02y=h-\dfrac{gh^2}{2v_0^2}y=h−2v02​gh2​.

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