A ball is dropped from a height of 45 m from the ground. The coefficient of restitution between the ball and the ground is 2/3. What is the distance travelled by the ball in 4th second of its motion. Assume negligible time is spent in rebounding :

Take $g=10m{s}^2$

The correct option is C 15 m
Let time taken by the ball to reach the ground be $t^{\prime }$ sec.

$\therefore$ Using $s=ut+\frac{1}{2}g(t^{\prime }{)}^2$

$45=0+\frac{1}{2}\times 10(t^{\prime }{)}^2$ $⟹t^{\prime }=3$ sec

Velocity of ball just before reaching the ground $u=\sqrt{2gh}$

$u=\sqrt{2\times 10\times 45}=30$ $m/s$

Velocity of ball just after collision $v=eu=\frac{2}{3}\times 30=20$ $m/s$

Distance travelled in 4th second i.e $t=1$ sec, $s^{\prime }=vt-\frac{1}{2}g{t}^2$

$s^{\prime }=20\times 1-\frac{1}{2}\times 10\times 1^2=15$ m