Question

A ball is dropped from a height. If it takes 0.200 s to cross the last 6.00 m before hitting the ground, find the height from which it was dropped. Take g = 10 m/s².

Answer 1

Given:
Distance travelled by the ball in 0.200 seconds = 6 m
Let:
Time, t = 0.200 s
Distance, s = 6 m
a = g = 10 m/s² (Acceleration due to gravity)
Using the equation of motion, we get:
$$\begin{gathered} s=ut+\frac{1}{2}a{t}^2 \\ 6=u\left(0.2\right)+\frac{1}{2}\times 10\times 0.04 \\ \Rightarrow u=\frac{5.8}{0.2}=29\mathrm{m}/\mathrm{s} \end{gathered}$$
Let h be the height from which the ball is dropped.
We have:
u = 0 and v = 29 m/s
Now,
$h=\frac{v^2-u^2}{2a}$
$\Rightarrow h=\frac{{29}^2-0^2}{2\times 10}=\frac{29\times 29}{20}=42.05\mathrm{m}$

∴ Total height = 42.05 + 6 = 48.05 m ≈ 48 m