Question

A ball is gently dropped from a height of 20m, if its velocity increases uniformly at the rate of 10m/s2 ,​

Answer 1

Answer:

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

Answer 2

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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

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