Distance travelled by the stone before striking the ground in last second
s = 44.1 m
t = 1s
And here the acceleration is taken as gravitational acceleration so,
a = g = 9.81 m/s2
Now, using the expression of distance travelled in time t with acceleration a and initial velocity is given by,
\(s = ut + \frac 12at^2 \quad........(i)\)
Where, s is distance travelled, t is time, a is acceleration, u is initial velocity.
So, substituting the values in expression (i) we will get,
\(44.1 = u(1) + \frac 12 (9.81) (1)^2\)
\(\Rightarrow 44.1 = u + 4.905\)
\(\Rightarrow u = 44.1 - 4.905 = 39.195\ m/s\)
So, initial velocity is 39.195 m/s.
Now, we know that relation between final velocity, acceleration and initial velocity is given by,
\(v^2 - u^2 = 2as \quad .........(ii)\)
Where, v is final velocity, u is initial velocity, a is acceleration and s is distance travelled.
Now, here v will be u of previous expression as the initial velocity becomes the final velocity of the stone just before striking the ground so it can be given as,
\(u=0 \)
\(v=39.195 m/s\)
\(a=g=9.81 m/s^2\)
So, using the above values we will find the distance travelled,
\((39.195)^2 - (0)^2 =2 (9.81)s\)
\(\Rightarrow \frac{1536.24}{2 \times 9.81} = s\)
\(\Rightarrow s= 78.3 m\)
Now, the total distance travelled by the stone before striking the ground is,
total distance = 78.3 + 44.1 = 122.4 m
Now, the total distance travelled by the stone is equal to the height of the cliff, so the height of the cliff is 122.4 m.
