Laws of Motion — Question 17

(i) 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,

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

Hence, maximum height attained = 122.5 m

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

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

Substituting the values in the formula, we get,

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

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