Question

A ball is gently dropped from a height of 20m. if its velocity increases uniformly at the rate of 10 m/s with what velocity will it strike the ground? after what time will is strike the ground?​

Answer 1

Answer:

v^2 = u^2 - 2gh

v^2 = 0^2m/s - 2*10m/s* 20m

v^2 = 400m/s^2

v = √400m/s^2

v = 20m/s

the ball hit the ground with the velocity of 20m/s

and time ,

t = v - u / g

[ from the equation. g = v - u / t ]

t = 20m/s - 0m/s / 10 m/s^2

t = 2s

2seconds will taken to strike the ground .

hope it helps you

Answer 2

Answer:

Initial velocity(u) = 0 m/s

Acceleration (a) = 10 m/s²

Time(s) = 20 m

We know,

v² = u² + 2as

Putting values:

v² = 0² + 2 (10 x 20)

v² = 400

v = 20 m/s

For time:

 v = u + at

t = v - u/a

t = (20-0)10

 = 20/10

 = 2

∴ Striking velocity = 20 m/s ← (Answer)

∴ Striking Time = 2 seconds  ← (Answer)