Physics, asked by Rihanashah, 1 month ago

A ball is gently dropped from a height of 20 m. If its 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?​

Answers

Answered by jaya13prabha
1

Answer:

Given, initial velocity of ball, u=0

Final velocity of ball, v=?

Distance through which the balls falls, s=20m

Acceleration a=10ms −2

Time of fall, t=?

As we know

2as = v2 – u2

v2 = 2as + u2

v2 = (2 x 10 x 20 ) + 0

v2 = 400

Final velocity of ball (v) = 20 ms-1

t = (v – u)/a

Time taken by the ball to strike (t) = (20 – 0)/10

t = 20/10

t = 2 seconds

The final velocity with which the ball will strike the ground is (v) = 20 ms-1

The time it takes to strike the ground (t) = 2 seconds

Answer is 20 m/s, 2 s

Answered by gurpreetkalra1981
1

Explanation:

initial velocity=0

distance=20

acceleration=10

final velocity= ?

by third equation of motion,

v²-u²=2as

v²-0²=2x10x20

v²-0=400

v²=400

v=√400

v=20

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