Question

A tennis ball is dropped from the top of a tall building of height H. It takes time t for the ball to reach the ground. At t/2 the ball has fallen through a distance:

[a] equal to H/2 [b] less than H/2 [c] more than H/2.

Answer 1

Answer 2

The distance fallen by the ball after time t/2 can be determined by analyzing the motion of the ball under gravity.

When an object is in free fall near the surface of the Earth, neglecting air resistance, it experiences a constant acceleration due to gravity, typically denoted by "g" and approximately equal to 9.8 meters per second squared (m/s^2). The acceleration due to gravity acts downward, causing the object to accelerate as it falls.

Let's consider the motion of the tennis ball at t/2:

1. The ball is initially at rest when dropped, so its initial velocity (u) is zero.

2. The acceleration due to gravity (a) is a constant and acts downward.

3. The time taken to reach t/2 is (t/2).

4. The distance fallen by the ball after time t/2 is the displacement (s) at t/2.

Using the kinematic equation:

s = ut + (1/2)at^2

Substituting the given values:

s = 0 + (1/2)gt^2/2

s = (1/4)gt^2

Since t is the total time it takes for the ball to reach the ground, we have t = 2t/2.

Substituting t = 2t/2 into the equation:

s = (1/4)g(2t/2)^2

s = (1/4)g(t^2/4)

s = (1/16)gt^2

Now, we can compare s to H/2:

If s = H/2, then (1/16)gt^2 = H/2.

Rearranging this equation:

gt^2 = 8H

t^2 = (8H)/g

t = √((8H)/g)

Therefore, the distance fallen by the ball after time t/2 is given by (1/16)gt^2. From the equation, we can see that (1/16)gt^2 is less than H/2. Thus, the correct answer is [b] less than H/2.