Question

13. A ball is thrown vertically upwards. It returns
6 s later. Calculate : (i) the greatest height reached
by the ball, and (ii) the initial velocity of the ball.
(Také g = 10 m s?) Ans. (i) 45 m, (ii) 30 m s-1
41​

Answer 1

Answer:

The ball is thrown up and it returns in 6sec,

Time of accent = Time of decent

So, time to reach the highest point =6/2=3s

By 2nd equation of motion

$s = ut + \frac{1}{2} at {}^{2}$

To calculate height, consider motion of the ball from highest point to the ground

$h = \frac{1}{2} gt {}^{2} \\ = 45m$

To calculate projection velocity, consider motion of the ball from projection to return to the point of projection.

$0 = ut - \frac{1}{2} gt {}^{2} \\ u = 30m/ \: s$