Question

A ball is gently dropped from a height of 45 M if its velocity increases in uniformly at the 10 metre per second square then what velocity is strikes the ground after what time it will strike​

Answer 1

Distance (s) = 45 m

Initial Velocity (u) = 0 m/s

Acceleration (a) = 10 m/s²

Then , finding the final velocity (v) :

The relation to be used :

v²-u² = 2as

v²-0² = 2×10×45

v² = 900

v = 30 m/s

Then , finding the time (t) :

The relation to be used :

v = u+at

30 = 0+10t

30/10 = t

3 sec = t

So ,

The striking velocity = 30 m/s

The striking time = 3 s

Answer 2

so the ball is increasing the velocity at the rate of 10$ms^{-2}$

so this would meant that it is it's acceleration

or

acceleration due to gravity "g"

so the velocity at the time when it just strike the ground would be

$v^2 - u^2 = 2gs$

$v^2-0 = 2×10×45$

$u^2$ is zero "0" before the time when the ball was to drop it was in rest

$v^2 = 900$

square root both side

$\sqrt{v^2} = \sqrt{900}$

$v = 30ms^{-1}$

so the velocity when it just reaches the ground would be $30ms^{-1}$

so the time taken to raech there

$a = \frac{v-u}{t}$

$10 = \frac{30-0}{t}$

$10t = 30$

$t = \frac{30}{10}$

$t = 3s$

hence time taken to reach there is 3 sec