Question

A ball is thrown vertically upwards attains a maximum height of 45m. What is the time after which velocity of the ball become equal to half the velocity of projection?

Answer 1

Let the velocity of projection be 'u'. Then 'v' = 0 (at top height)


By equation of motion, v²=u²+2as
i.e.,
v2 = u2 - 2gh (Since here s = h and when ball is thrown accealaration is taken as -g, i.e., a=-g)

u = 2gh−−−√

u = 30 m/s

now, time, when v' = 15 m/s

from first equation of motion: v= u+at
Here a=-g
So,
v = u -gt
15 = 30 - 10t

10t = 15

t = 1.5 s

So after 1.5 seconds velocity becomes half.