Question

A stone is thrown vertically upward with a speed of 10.0 m/s from the edge of a cliff 65.0m high. (a) how much later does it reach the bottom? (b) what is its speed at the bottom? (just before it hits the ground). (c) what total distance did it travel? (d) what was the total displacement?

Answer 1

This problem can be divided in to 2 parts

First part in which stone moves up and second part in which stone moves down.

Let the time for travel in upper direction be $t_1$ and time for travel in down direction be $t_2$

We have v = u+at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

  Up direction

               0 = 10 - 9.81*$t_1$

                $t_1$ = 1.02 seconds

We also have $s=ut+\frac{1}{2} at^2$, s is the distance traveled.

                             s = 10*1.02-$\frac{1}{2} *9.81*1.02^2 = 5.09 m$

Now considering the downward motion.

                  displacement = 5.09+65=70.09 m

                 $s=ut+\frac{1}{2} at^2$

                 $70.09 = 0*t+0.5*9.81*t_2^2\\ \\ t_2=3.78seconds$

a) So total time taken = 3.78+1.02 = 4.8 seconds

b) In case of downward motion v = u+at

          u=0, a = g, t = 3.78 seconds

  So, Velocity on reaching ground = 0+9.81*3.78 = 37.08 m/s

c) Total distance traveled = 5.09+70.09 = 75.18 m

d)Total displacement is height of cliff, which is equal to 65 m