Question

A ball is dropped from a height. It takes 0.200 s to cross the last 6.00 before hiting the ground, find the height from which it was dropped. Take $g=10m/s^2$

Answer 1

For last 6 m Distance travelled,

S = 6 m,

u =?, t = 0.2 sec, a=g = 9.8 $m/s^2$

S = $ut+\frac{1}{2}a{t}^2$

$6=u\left(0.2\right)+4.9\times 0.04$

$\Rightarrow u=\frac{5.8}{0.2}=29m/s$

For distance x, u = 0, v = 29 m/s

a = g = 9.8 $m/s^2$

$\therefore S_2=\frac{v^2-u^2}{2a}$

= $\frac{{29}^2-0^2}{2}\times 9.8=\frac{29\times 29}{19.6}$

= 42.5 m

Total distance = 42.5 + 6 = 18.5 m =48 m