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?
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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
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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