A rubber ball is dropped from a height of 5m on a planet where the acceleration due to gravity is not known. On bouncing, it rises to 1.8m . The ball loses its velocity on bouncing by a factor of

The correct option is 3/5.

Let us solve the problem step-by-step:

Step 1: Understand the motion of the ball before bouncing.
When the rubber ball is dropped from a height, it accelerates downwards due to gravity. Let the acceleration due to gravity on the planet be $g$ and the initial dropping height be:
$h_1=5\text{m}$
Using the equation of motion, the velocity of the ball $v_1$ just before it hits the ground is given by:
$v_1^2=2g{h}_1$
Taking the square root on both sides:
$v_1=\sqrt{2g{h}_1}$

Step 2: Understand the motion of the ball after bouncing.
On bouncing, the ball rises to a height:
$h_2=1.8\text{m}$
Let $v_2$ be the velocity of the ball immediately after bouncing as it starts to rise. At its maximum height of 1.8 m, its velocity becomes zero. Using the equation of motion for this upward journey:
$v_2^2=2g{h}_2$
Taking the square root on both sides:
$v_2=\sqrt{2g{h}_2}$

Step 3: Calculate the ratio of the velocities.
To find the factor of velocity after bouncing compared to the velocity before bouncing, we take the ratio of $v_2$ to $v_1$:
$\frac{v_2}{v_1}=\frac{\sqrt{2g{h}_2}}{\sqrt{2g{h}_1}}$
Simplifying the expression by cancelling the common terms:
$\frac{v_2}{v_1}=\sqrt{\frac{h_2}{h_1}}$

Step 4: Substitute the given values.
Substitute the values of the heights into the ratio:
$\frac{v_2}{v_1}=\sqrt{\frac{1.8}{5}}$
To simplify, multiply the numerator and the denominator inside the square root by 10:
$\frac{v_2}{v_1}=\sqrt{\frac{18}{50}}$
Simplify the fraction by dividing both numbers by 2:
$\frac{v_2}{v_1}=\sqrt{\frac{9}{25}}$
Taking the square root yields:
$\frac{v_2}{v_1}=\frac{3}{5}$

Thus, the ball's velocity after bouncing changes by a factor of 3/5.