Question

A man is standing on the edge of a 20.0 m high cliff. He throws a rock vertically with aninitial velocity of 10.0 m/s.a) How high does the rock go? (Remember that at its max height v = 0 m/s)b) How long does it take to reach its max height?

Answer 1

HGiven,

The height of the cliff, h- 20.0 m

The initial velocity of the rock, u=10.0 m/s

The final velocity of the rock, v=0 m/s

The acceleration acting on the rock is g=-9.8 m/s. The negative sign indicates that it is in the direction opposite to that of the initial velocity, i.e., downward direction.

a) From the equation of the motion, we have

`v^2-u^2=2gs`

Where s is the maximum height reached by the rock

On substituting the known values,

`\begin{gathered} 0-10^2=2*-9.8* s \\ \Rightarrow s=(-10^2)/(2*-9.8) \\ s=5.1\text{ m} \end{gathered}`

Thus the maximum height reached by the rock above the cliff is 5.1 m

The maximum height reached by the rock from the ground is,