Question

A body projected upwards at same height from ground at t = 1s and t=9s. Find the maximum height attained
by the ball.
Take g = 10m/s^2​

Answer 1

Answer :-

h = 125 m

Given :-

A ball is projected upwards at same height at t = 1s and t = 9s.

To find :-

maximum height attained by the ball.

Solution:-

Let ball attained the height h metre.

• By using equation of motion.

→$h = ut - \dfrac{1}{2}gt^2$

$h = u \times t - \dfrac{1}{2}\times 1 \times (1)^2$

→$h = u - 5 -------1)$

→$h = u \times 9 - \dfrac{1}{2}\times 10 \times(9)^2$

→$h = 9u - 405-------2)$

• Both the height is same.

→$u - 5 = 9u -405$

→$u -9u = -405 +5$

→$-8u = -400$

→$u = \dfrac{-400}{-8}$

→$u = 50 m /s$

• At maximum height v = 0

→$2gh = v^2 -u^2$

→$2 \times -10 \times h = (0)^2 - (50)^2$

→$-20h = -2500$

→$h = \dfrac{-2500}{-20}$

→$h = 125 m$

hence,

Maximum height attained by body will be 125m.