Question

A ball is gently dropped from a height of 45 m. If it's velocity increases uniformly at the rate of 10 m/s^2 , with what velocity will it strike the ground? After what time will it strike the ground?​

Answer 1

Answer:

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

Answer 2

Explanation:

initial velocity u= 0

distance covered s = 45 meter

final velocity = v

tim=t

hope it helps