Question

A ball is projected vertically upward with a velocity u from ground. If t is the time taken by ball to reach the maximum height (H), then what will be the height of the ball from the ground after 7t/5?

Answer 1

The trick to answering this is that after time 't' the ball falls freely. So, the height of the ball at time 7t/5 is equal to the height of the ball at 7t/5-t= 2t/5.

Answer 2

Answer: $\frac{21}{25}H$

Explanation:

Time taken by ball to reach maximum height = time taken by ball to hit ground.

The velocity at the maximum height = 0

hence, $H=\frac {1}{2}gt^2$

where g is the acceleration due to gravity, t is the time taken by ball to cover a distance H

At time 7t/5, the ball would be in downward motion. we need to find the distance covered in 7t/5-t =2t/5 because time t is taken in the upward flight.

After 2t/5 time from the maximum height, distance covered,

$s=\frac{1}{2}g(2t/5)^2=\frac{2}{25}gt^2=\frac{4}{25}H$

Hence, the height of the ball from the ground,

$h'=H-\frac{4}{25}H=\frac{21}{25}H$