Laws of Motion — Question 15

(i) From the equation of motion,

h = ut + $\dfrac{1}{2}$ gt²

Given,

g = 10 m s^-2

h = 80 m

u = 0

Substituting the values in the formula, we get,

$80 = (0 \times \text t) + (\dfrac{1}{2} \times 10 \times \text t^2) \\[0.5em] 80 = 0 + (5 \times \text t^2) \\[0.5em] 80 = 5 \times \text t^2 \\[0.5em] \text t^2 = \dfrac{80}{5} \\[0.5em] \text t^2 = 16 \\[0.5em] \text t = 4\ \text s \\[0.5em]$

Hence, time taken = 4 s

(b) From the equation of motion,

v² = u² + 2gh

u = 0

g = 10 m s^-2

h = 80 m

Substituting the values in the formula, we get,

$\text v^2 = 0^2 + (2 \times 10 \times 80) \\[0.5em] \text v^2 = 1600 \\[0.5em] \Rightarrow \text v = 40 \text{ m s} ^{-1}\\[0.5em]$

Hence, final velocity of the stone on reaching the ground = 40 m s^-1