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The Equation for the Path of Projectile

A projectile ...

Question

A projectile is fired with an initial speed of ${500 m s}^{- 1}$ 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?

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Solution

Given,

Initial speed $= 500 m / s$

$h e i g h t = 19.6 m$

So, the equation of motion in y-diraction:

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

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

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

$t^{2} = 4$

$t = 2 s$

Thus the equation of motion is x-direction:

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

$s = 500 \times 2 = 1000 m$

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

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

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