Question

A ball is kicked off a 75 m tall cliff with a horizontal velocity of 5 m/s. How long till it reaches the ground? How far will the ball land from the base of the cliff?

Answer 1

Let, time taken is t.

So,

$s=ut + \dfrac{gt^2}{2}\\\\75 = 0 + \dfrac{10\times t^2}{2}\\\\150=10\times t^2\\\\t = \sqrt{15}\\\\t = 3.873\ s$ ( Here, us is vertical component of initial velocity )

Now, distance covered from the base of the cliff is :

$D=5\times 3.873\ m\\\\D = 19.365\ m$

Therefore, time taken and distance covered is 3.873 s and 19.365 m.

Hence, this is the required solution.