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 [tex]t_1[/tex] and time for travel in down direction be [tex]t_2[/tex]
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*[tex]t_1[/tex]
[tex]t_1[/tex] = 1.02 seconds
We also have [tex]s=ut+\frac{1}{2} at^2[/tex], s is the distance traveled.
s = 10*1.02-[tex]\frac{1}{2} *9.81*1.02^2 = 5.09 m[/tex]
Now considering the downward motion.
displacement = 5.09+65=70.09 m
[tex]s=ut+\frac{1}{2} at^2[/tex]
[tex]70.09 = 0*t+0.5*9.81*t_2^2\\ \\ t_2=3.78seconds[/tex]
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
