Question

A ball is dropped from a height of 5 m onto a sandy floor and penetrates the sand up to 10 cm before coming to rest. Find the retardation of the ball is sand assuming it to be uniform.

Answer 1

A ball is dropped from a height of 5 m (s) above the sand level.
The same ball penetrates the sand up to 10 cm (s_s) before coming to rest.
Initial velocity of the ball, u = 0 And,
a = g = 9.8 m/s² (Acceleration due to gravity)
Using the equation of motion, we get : 

\[s = ut + \frac{1}{2}a t^2\]

\[\Rightarrow 5 = 0 + \frac{1}{2}\left( 9 . 8 \right) t^2 \]

\[ \Rightarrow t^2 = \frac{5}{4 . 9} = 1 . 02\]

\[ \Rightarrow t = 1 . 01 s\]

Thus, the time taken by the ball to cover the distance of 5 m is 1.01 seconds.
Velocity of the ball after 1.01 s:
 v = u + at
 ⇒ v = 9.8 × 1.01 = 9.89 m/s     
Hence, for the motion of the ball in the sand, the initial velocity u₂ should be 9.89 m/s and the final velocity v₂ should be 0.
s_s = 10 cm = 0.1 m
Again using the equation of motion, we get:

\[a_s = \frac{v_2^2 - u_2^2}{2 s_s} = \frac{0 - \left( 9 . 89 \right)^2}{2 \times 0 . 1}\]

\[ \Rightarrow a_s \approx - 490 \text{ m } / s^2\]

Hence, the sand offers the retardation of 490 m/s².