Class 11 NEET Physics Motion in a Straight Line Questions

We know that when a ball is dropped from height $h$ , it strikes the surface with speed $v=\sqrt{\frac{2gh}{1}}$ It is given that the ball is dropped from height $5 \text{ m}$ . Substituting this value and value of acceleration due to gravity in the formula we get $v=\sqrt{2 \times 10 \times 5}$ , $v=\sqrt{100}$ , $v=10 \text{ m/s}$ We also know that when a ball rebounds with speed $u$ , the height it reaches is given by $H=\frac{u^2}{2g}$ It is given that the ball reaches height $1.8 \text{ m}$ on rebound. Substituting this value and value of acceleration due to gravity in the formula, we get $1.8=\frac{u^2}{2 \times 10}$ $u^2=36$ , $u=6 \text{ m/s}$Hence the ball loses its velocity by $4 \text{ m/s}$ on rebounding. The fractional loss can be calculated by dividing the loss in velocity with initial velocity, i.e. velocity with which it strikes the floor i.e $10 \text{ m/s}$ Hence, fractional loss in velocity $=\frac{4}{10}$, fractional loss in velocity $=\frac{2}{5}$.