Question

A ball is gently dropped from a high of 20m. If it's velocity increases uniformly at the rate of 10 ms -2 what with velocity will it strike the ground? After what time will it strike the ground

Answer 1

Answer:

if velocity or acceleration is 10ms—² so the velocity it will strike the ground is

$10 \times 20 = 200ms {2}$

Answer 2

Answer:

Explanation:

Initial Velocity= 0 m/s = u      time = t

acceleration = 10 ms-2 = a

distance = 20 m = s

s=u t + 1/2 a t 2   [t 2 is time square ]

20 m =0 + 5 ms-2*t 2

20 m /5 ms-2 = t 2

4 s 2 = t 2

2 s =t  

time = 2 s

final velocity = v

a = [ v+u] / t

10 ms-2 =[ v + o ] / 2 s

10 ms-2 * 2 s = v

20 m / s = v

final velocity = 20 m / s