A ball is thrown vertically upwards with an initial velocity of 49 m s^-1. Calculate: (i) the maximum height attained, (ii) the time taken by it before it reaches the ground again. (Take g = 9.8 m s^-2)

From the equation of motion,

v² = u² - 2gh (a = -g , as the movement is against gravity)

u = 49 m s^-1

v = 0

g = 9.8 m s^-2

Substituting the values in the formula, we get,

$0^2 = 49^2 - 2 \times 9.8 \times h \\[0.5em] 2401 = 19.6 \times h \\[0.5em] \Rightarrow h = \dfrac{2401}{19.6} \\[0.5em] \Rightarrow h = 122.5 m\\[0.5em]$

Hence, maximum height attained = 122.5 m

(ii) As we know, total time of journey is,

t = $\dfrac{2u}{g}$

Substituting the values in the formula, we get,

$t = \dfrac{2 \times 49}{9.8} \\[0.5em] t = \dfrac{98}{9.8} \\[0.5em] t = 10 s \\[0.5em]$

Hence, the time taken by the ball before it reaches the ground again = 10 s