A ball is gently dropped from a height of 20m. If its velocity increase uniformly at the rate of 10 m/s, with what velocity will it strike the ground ? After what time will it strike the ground ?
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Answers
Explanation:
Answer:
Let us assume, the final velocity with which the ball will strike the ground be ‘v’ and the time it takes to strike the ground be ‘t’.
Given parameters
Initial Velocity of the ball (u) = 0
Distance or height of fall (s) = 20 m
Downward acceleration (a) = 10 m s-2
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
Given parameters
Initial Velocity of the ball (u) = 0
Distance or height of fall (s) = 20 m
Downward acceleration (a) = 10 m s-2
As we know
2as = v^2 – u^2
v^2 = 2as + u^2
v^2 = (2 x 10 x 20 ) + 0
v^2 = 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