Two stones are thrown simultaneously, one straight upward from the base of a cli

Question

Two stones are thrown simultaneously, one straight upward from the base of a cliff and the other straight downward from the top of the cliff. The height of the cliff is 6.63m. The stones are thrown with the same speed of 8.49m/s. Find the location (above the base of the cliff) of the point where the stones cross paths. Two stones are thrown simultaneously, one straight upward from the base of a cliff and the other straight downward from the top of the cliff. The height of the cliff is 6.63m. The stones are thrown with the same speed of 8.49m/s. Find the location (above the base of the cliff) of the point where the stones cross paths.

Explanation / Answer

Lets start with position formula: x-x0 = v0t + 1/2 a t^2
For the stone thrown from the top of the cliff we have:
x1 - ( 6.63) = (-8.49)t + (1/2)(-9.8)t^2
x1 = (1/2)(-9.8)t^2 + (-8.49)t + (6.63)
9.8 is g. Lets assume anything upward is positive and anything going down is negative.
For the stone thrown from the base we have:
x2 - (0) = (8.49)t + (1/2)(-9.8)t^2

Now we want to find the time which they met. We know when they met the position for both stones are the same (other wise they wouldn't meet :) ).
x1 = x2
(1/2)(-9.8)t^2 + (-8.49)t + (6.63) = (8.49)t + (1/2)(-9.8)t^2
t=0.39 s

x1 = (1/2)(-9.8)t^2 + (-8.49)t + (6.63)
x1=3.64

x2 - (0) = (8.49)t + (1/2)(-9.8)t^2
x2=2.988
now you can solve for "t", which tells you how long tool the stones to travel to meet.
Note: it would take less than a second for one stone to hit the bottom of the cliff since the cliff is only 6.63m, so that should be a clue that you should expect a very small number for "t" from the above formula.

now you can plug-in the number you've got from t in the same formula (x-x0 = v0t + 1/2 a t^2) to get "x",
x - 6.63=9.17 x 0.3 +1/2 (9.17/0.3)0.3 x 0.3
=10.7565 m

Calculation may be wrong but concept is right

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