Question

A rock is dropped from a vertical cliff. The rock takes 3.00 s to reach the ground below the cliff. A second rock is thrown vertically from the cliff. It takes 2.00 s to reach the ground below the cliff from the time it is released. With what velocity was the second rock released?

Answer 1

Answer:

Velocity v= 12.25 $\frac{m}{s}$

Explanation:

The first rock dropped give the distance Y in meters

$Y_{f}=Y_{o}+v_{o}*t +\frac{1}{2}*a*t^{2}\\ Y_{f}=\frac{1}{2}* 9.8 \frac{m}{s^{2} }* 3^{2} \\Y_{f}=44.1 m$

Now the motion of the second rock the time change so to know the velocity

$Y_{f}= Y_{o} +v_{o}*t +\frac{1}{2}*a *t^{2} \\v_{o}*t= -Y_{f} +\frac{1}{2} *a*t^{2} \\v_{o} =\frac{-44.1 +0.5 * 9.8* s^{2} }{2} \\v_{o}= 12.25 \frac{m}{s}$