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. (Take g = 10 m s^-2 )

From the equation of motion,

h = ut + $\dfrac{1}{2}$ gt²

(we consider motion of ball from highest point to the ground)

Since, total time for journey of the ball in air = 6 s.
Therefore, time for reaching maximum height = $\dfrac{6}{2} = 3$ s

g = 10 m s^-2

u = 0

Substituting the values in the formula, we get,

$h = (0 \times 3) + (\dfrac{1}{2} \times 10 \times 3^2) \\[0.5em] h = 0 + (5 \times 9) \\[0.5em] h = 45 m \\[0.5em]$

Hence, greatest height reached by the ball = 45 m

(ii) From the equation of motion,

v² = u² - 2gs

Substituting the values in the formula, we get,

$0^2 = u^2 - (2 \times 10 \times 45) \\[0.5em] u^2 = (2 \times 10 \times 45) \\[0.5em] u^2 = 900 \\[0.5em] u = 30 \text {m s}^{-1} \\[0.5em]$

Hence, the initial velocity of the ball = 30 m s^-1