Question

A ball is gently dropped from a height of 20m if its velocity inof creases uniformly at the rate of 10m/s2, with what velocitywill it strikes the ground

Answer 1

Given that, height = 20 m

$\sf\implies s = 20 m$

And, it accelerates uniformly at the rate of 10 m/s²

$\sf\implies a = 10 m s^{-2}$

Now, since the ball is gently dropped, it's initial velocity, or u would be 0 m/s

So, we have s = 20 m

u = 0 m/s

a = 10 m/s²

Let Final velocity = v

Now, substituting these values in third equation of motion, we get :-

$\sf 2as = v^2 - u^2$

$\implies\sf 2(10)(20) = v^2 - 0$

$\implies\sf 400 = v^2$

$\implies\sf v = \sqrt{400}$

$\implies\sf v = 20 m s^{-1}$

Hence, the ball will strike the ground with a velocity of 20m/s.

Answer 2

u = 0 (since dropped)

v = ?

a = 10

s = 20

using the third equation of motion, we get

v^2 = 0 + 2×10×20

v^2 = 400

v = 20 m/s