A ball is released from a height of 20m calculate
The time it takes to fall
The velocity with which it hits the ground
The Correct Answer and Explanation is :
To calculate the time it takes for the ball to fall and the velocity with which it hits the ground, we’ll use the basic principles of motion under gravity. In this case, we are assuming the motion occurs in a vacuum (no air resistance), and the only force acting on the ball is gravity.
Given:
- Height (initial position) = 20 m
- Acceleration due to gravity ((g)) = 9.8 m/s² (downward)
- Initial velocity ((u)) = 0 m/s (since the ball is released and not thrown)
1. Time it takes to fall
The ball is falling from rest, so we can use the second equation of motion:
[
s = ut + \frac{1}{2}gt^2
]
Where:
- (s) is the displacement (20 m, downward)
- (u) is the initial velocity (0 m/s)
- (t) is the time taken to fall
- (g) is the acceleration due to gravity (9.8 m/s²)
Substituting the known values:
[
20 = 0 \times t + \frac{1}{2} \times 9.8 \times t^2
]
Simplifying:
[
20 = 4.9t^2
]
Solving for (t^2):
[
t^2 = \frac{20}{4.9} \approx 4.08
]
Taking the square root:
[
t \approx \sqrt{4.08} \approx 2.02 \, \text{seconds}
]
2. Velocity with which it hits the ground
To find the velocity with which the ball hits the ground, we use the first equation of motion:
[
v = u + gt
]
Where:
- (v) is the final velocity (which we need to find)
- (u) is the initial velocity (0 m/s)
- (g) is the acceleration due to gravity (9.8 m/s²)
- (t) is the time taken to fall (2.02 s)
Substituting the values:
[
v = 0 + 9.8 \times 2.02
]
[
v \approx 19.8 \, \text{m/s}
]
Conclusion:
- The time it takes for the ball to fall is approximately 2.02 seconds.
- The velocity with which it hits the ground is approximately 19.8 m/s downward.
Explanation:
The ball is subject only to gravitational force, which causes it to accelerate at a constant rate ((9.8 \, \text{m/s}^2)) downwards. The equations of motion allow us to calculate the time it takes for the ball to fall from rest and the final velocity just before it hits the ground. The calculations use basic kinematic formulas, which are derived from the principles of motion under constant acceleration.