Laws of Motion — Question 9

(a) As we know from the equation of motion;

s = ut + $\dfrac{1}{2}$gt²

where, s = height

Given,

t = 3 s

g = 9.8 m s^-2

initial velocity (u) = 0

Substituting the values in the formula above, we get,

$\text S = (0 \times t) + (\dfrac{1}{2} \times 9.8 \times 3^2) \\[0.5em] \text S = 0 + (\dfrac{1}{2} \times 9.8 \times 9) \\[0.5em] \text S = 44.1\ \text m \\[0.5em]$

Hence, the height from which the ball was released = 44.1 m

(b) From the equation of motion,

v² = u² - 2gs

where, v = final velocity

Substituting the values in the formula, we get,

$\text v^2 = 0^2 - (2 \times 9.8 \times 44.1) \\[0.5em] \text v^2 = 2 \times 9.8 \times 44.1 \\[0.5em] \Rightarrow \text v^2 = 864.36 \\[0.5em] \Rightarrow \text v = 29.4 \text{ m s}^{-1} \\[0.5em]$

Hence, velocity with which the ball strikes the ground = 29.4 m s^-1