Question

A projectile is fired with an initial speed of ${500ms}^{-1}$ horizontally from the top of a cliff of height 19.6m. At what distance from the foot of the cliff does it strike the ground?

Answer 1

Given,

Initial speed $=500m/s$

$height=19.6m$

So, the equation of motion in y-diraction:

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

$\Rightarrow 19.6=0+\frac{1}{2}\times 9.8\times t^2$

$\Rightarrow \frac{19.6}{4.9}=t^2$

$t^2=4$

$t=2s$

Thus the equation of motion is x-direction:

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

$s=500\times 2=1000m$

Where, $u=5000m/s,t=2s,a=0$

So, At $1000m$ from the foot of the cliff it strikes the ground.