A child drops a ball from a height of 10m. Assume that its velocity increases uniformly at the rate of 10m/s. Find (i) the velocity with which the ball strikes the ground (ii) the time taken by the ball to reach the ground.
Answers
Correct Question :
›»› A child drops a ball from a height of 10m. Assume that its velocity increases uniformly at the rate of 10m/s². Find (i) the velocity with which the ball strikes the ground (ii) the time taken by the ball to reach the ground.
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Answer :
›»› The velocity with which the ball strikes the ground = 10√2 m/s
›»› The time taken by the ball to reach the ground = 1.41 sec
Given :
- Initial velocity of ball (u) = 0 m/s
- A child drops a ball from a height (s) = 10 m
- Acceleration (a) = 10 m/s²
To Find :
- Final velocity with which the ball hits the ground = ?
- Time taken by the ball to reach ground = ?
Required Solution :
→ Initial velocity (u) = 0 m/s (because the ball has been dropped freely).
❶ The velocity with which the ball strikes the ground
From third equation of motion
⇛ v² = u² + 2as
⇛ v² = 0² + 2 × 10 × 10
⇛ v² = 0 + 2 × 10 × 10
⇛ v² = 0 + 20 × 10
⇛ v² = 0 + 200
⇛ v² = 200
⇛ v = √200
⇛ v = √(100 × 2)
⇛ v = √(10² × 2)
⇛ v = 10√2 m/s
║Hence, the velocity of ball with which the ball strikes the ground is 10√2 m/s.║
Now, we have three elements that used in formula, Final velocity, Initial velocity and acceleration,
- Final velocity (v) = 10√2 m/s
- Initial velocity (u) = 0 m/s
- Acceleration (a) = 10 m/s
And we need to find Time taken by the ball to reach the ground.
❷ Time taken by the ball to reach the ground.
From first equation of motion
⇛ v = u + at
⇛ 10√2 = 0 + 10 × t
⇛ 10√2 = 0 + 10t
⇛ 10√2 = 10t
⇛ t = (10√2)/10
⇛ t = √2
⇛ t = 1.41 sec
║Hence, the Time taken by the ball to reach ground is 1.41 sec.║