Answer:
[tex]t=\sqrt{2h/g}-(1/g)*(\sqrt{v_{o}^2+2gh}-v_{o})[/tex]
Explanation:
First person:
[tex]y(t)=y_{o}-v_{o}t-1/2*g*t^{2}[/tex]
[tex]v_{o}=0[/tex] the rock is dropped
[tex]y_{o}=h[/tex]
[tex]y(t)=h-1/2*g*t^{2}[/tex]
after t1 seconds it hit the ground, y(t)=0
[tex]0=h-1/2*g*t_{1}^{2}[/tex]
[tex]t_{1}=\sqrt{2h/g}[/tex]
Second person:
[tex]y(t)=y_{o}-v_{o}t-1/2*g*t^{2}[/tex]
[tex]v_{o}[/tex] the rock has a initial downward speed
[tex]y_{o}=h[/tex]
[tex]y(t)=h-v_{o}t-1/2*g*t^{2}[/tex]
after t2 seconds it hit the ground, y(t)=0
[tex]0=h-v_{o}t_{2}-1/2*g*t_{2}^{2}[/tex]
[tex]g*t_{2}^{2}+2v_{o}t_{2}-2h=0[/tex]
[tex]t_{2}=(1/2g)*(-2v_{o}+\sqrt{4v_{o}^2+8gh})[/tex]
the time t when the second person throws the rock after the first person release the rock is:
t=t1-t2
[tex]t=\sqrt{2h/g}-(1/g)*(\sqrt{v_{o}^2+2gh}-v_{o})[/tex]
