Question

At t=0 a ball is thrown straight upward from the ground level at speed Vo. At the same instant, a distance D above, a ball is thrown straight downward, also at speed Vo.

Where do the balls collide in terms of D, Vo, and g?

Answer 1

\[h_1=v_0t-\frac{ 1 }{ 2}g t^2\]\[h_2=D-(v_0t+\frac{ 1 }{ 2}g t^2)\]\[h_1=h_2 \rightarrow 2v_0t=D \rightarrow t=\frac{ D }{ 2v_0 }\]\[h_1=v_0\left( \frac{ D }{ 2v_0 } \right)-\frac{ 1 }{ 2}g \left( \frac{ D }{ 2v_0 } \right)^2=\frac{ D }{ 2 }-\frac{ 1 }{ 8}g \frac{ D^2 }{ v_0^2 }\]If you replace in h2 formula, you have to obtain the same result:\[h_2=D-v_0\frac{ D }{ 2v_0}-\frac{ 1 }{ 8}g \frac{ D^2 }{ v_0^2 }=\frac{ D }{ 2 }-\frac{ 1 }{ 8}g \frac{ D^2 }{ v_0^2 }\]