We will take velocity and distance as positive upwards, and distance = 0 at the bottom of the cliff.
Use
Distance = Vi*t-(1/2)gt²
For the stone thrown upwards,
X1(t)=9.33t-(1/2)gt^2
For the stone thrown downwards,
X2(t)=5.28-9.33t-(1/2)gt^2
When they meet, X1(t)=X2(t)
and solve for t.
For checking your answer, 0.25<t<0.3 s.
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 5.28 m. The stones are thrown with the same speed of 9.33 m/s. Find the location (above the base of the cliff) of the point where the stones cross paths.
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