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:
$\mathrm{s}=ut+\frac{1}{2}a{t}^2$
$$\begin{gathered} \Rightarrow 5=0+\frac{1}{2}\left(9.8\right)t^2 \\ \Rightarrow t^2=\frac{5}{4.9}=1.02 \\ \Rightarrow t=1.01\mathrm{s} \end{gathered}$$

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:
$$\begin{gathered} a_{\mathrm{s}}=\frac{v_2^2-u_2^2}{2{\mathrm{s}}_s}=\frac{0-{\left(9.89\right)}^2}{2\times 0.1} \\ \Rightarrow a_{\mathrm{s}}\approx -490\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$
Hence, the sand offers the retardation of 490 m/s².