Question

A ball is dropped from a building of height h. Derive an equation for its velocity just before it hits the ground. Assume no air resistance.

Answer 1

Answer:

The final velocity of the ball just hits the ground is $\sqrt{2gh}$.

Explanation:

A ball is dropped from the height h. As the ball is dropped means there is no initial velocity given to the ball.

GIVEN:

The distance traveled by the ball is equal to h.

Initial velocity of the ball is zero.

Concept:

Two type of force comes into play in dropping of ball.

1. Gravitational force

Due to gravitational force ball forced to fall downward towards the surface.

2. Air resistance

Air develops drag force on the ball in opposite direction of the gravitational force.

Assumption:

There is no air resistance in falling of ball. So, there will be no drag force on the ball.

Calculation:

It is given that a ball is dropped from a building of height h.

The Newton’s equations of motion are as follows:

$V^{2}-u^{2}=2as$

$s=ut+\frac{1}{2}at^{2}$

$v=u+at$

STEP 1

Newton’s equation of motion is expressed as follows:

$V^{2}-u^{2}=2as$

Here, V is final velocity of the ball just before the ball hits the ground, a is acceleration due to gravity, u is the initial velocity of the ball and s is the distance traveled by the ball.

Here, initial velocity is zero. So, u = 0 m/s.

Vertical distance traveled by the ball is h. So, s= h.

Only gravitational force is responsible for motion as well as for acceleration. Thus, the acceleration due to gravity acts on the ball. Acceleration due to gravity is denoted by small g.

Acceleration due to gravity is constant for different planate. The acceleration due to gravity for the earth is 9.8 m/s². Here, no planet is defined. So the acceleration due to gravity is a = g.

Substitute 0 for m/s, g for a and h for s in above equation as follows:

$V^{2}-u^{2}=2as$

$V^{2}-(0)^{2}=2gh$

$V^{2}=2gh$

$v=\sqrt{2gh}$

Thus, the final velocity of the ball just hits the ground is $\sqrt{2gh}$.