Answer:
She stone hit ground 42.86 m far from base of cliff.
Explanation:
Initial height from ground = 40 meter.
We have equation of motion , [tex]s= ut+\frac{1}{2} at^2[/tex], s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In this the velocity of body in vertical direction = 0 m/s, acceleration = 9.8 [tex]m/s^2[/tex], we need to calculate time when s = 40 meter.
Substituting
[tex]40=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 2.86 seconds[/tex]
So it will take 2.86 seconds to reach ground.
So, stone travels horizontally at 15 m/s for 2.86 seconds.
Distance travelled = 15 x 2.86 = 42.86 m
So, the stone hit ground 42.86 m far from base of cliff.
