A ball is dropped from the top of a 54m high cliff. At the same time, a carefully aimed stone is thrown straight up from the bottom of the cliff with a speed of 27m/s^2 . The stone and ball collide part way up.

Question

anonymous · Answer

How far above the base of the cliff does this happen?

anonymous · Answer

If I were to make an equation for the height of the dropped ball as a function of time, it would be
$$ h(t) = H -\frac{1}{2} gt^2 $$
where H is the height of the cliff.  Can you make an equation for the height of the ball that was thrown upward?

anonymous · Answer

i have 
$$y=\frac{ v ^{2}-v _{0}^{2} }{ 2g }$$

$$v ^{2}=27m/s^2$$

anonymous · Answer

wait, i got that wrong. hold on

anonymous · Answer

As a function of time, the height from the bottom of the cliff can be written
$$ h(t) = v_0 t - \frac{1}{2}gt^2 $$

anonymous · Answer

Set those two equal to one another.  What do you get (solving for t) ?

anonymous · Answer

i got 2 seconds

anonymous · Answer

Symbolically, 
$$t = \frac{H}{v_0} $$
Plug that into either of the two height formulas, and what do you get?

anonymous · Answer

so they collide at 34.4 meters

anonymous · Answer

$$h(t) = H - \frac{1}{2}gt^2 = H - \frac{gH^2}{2v_0^2} $$
or
$$h(t) = v_0t - \frac{1}{2}gt^2 = H - \frac{gH^2}{2v_0^2} $$

anonymous · Answer

sweet. thank you for your help again, you are the man.
i have to go to class now but ill see you around here again.
take care

anonymous · Answer

No problem.  Likewise.