Question

A ball of mass 200 gram fall from a height of 5 metre what is its kinetic energy when it reaches the ground (g= 9.8m/sec)

Answer 1

Given :

• Mass of the ball (m) = 200 g
• Height (h) = 5 m
• Acceleration due to gravity (g) = 9.8 m/s²

To Find :

• kinetic energy of the ball when it reaches the ground = ?

Solution :

First convert the unit of the mass of ball from gram to kilogram :

→ Mass of ball = 200 g

→ Mass of ball = (200) ÷ (1000)

→ Mass of ball = 2 ÷ 10

→ Mass of ball = 0.2 kg

• Hence,the mass of the ball is 0.2 kg.

According to the Question Now :

• Finding potential energy at higest point :

→ Potential energy = mass × Acceleration due to gravity × Height

→ Potential energy = 0.2 × 9.8 × 5

→ Potential energy = 1.96 × 5

→ Potential energy = 9.8 J

• Hence,the potential energy of the ball is 9.8 J.

As we know that, the Potential energy at the highest point is equal to the Kinetic energy at the lowest point.

• Hence,from the above statement we can conclude that the Kinetic energy of the ball when it reaches the ground will be 9.8 J.

Answer 2

Given:

Mass of the ball: 200g or 2/10 Kg

g= 9.8 m/sec square

Solution:

Kinetic energy= $\frac{1}{2}mv^2$

By equation of motion,

$v^2= u^2+2as \\\\ v^2= 0+ 2 \times 9.8 \times 5 = 98$

Now, put the values in Kinetic energy formula $= \frac{1}{2}mv^2 \\\\= \frac{1}{2}\times\frac{2}{10} \times 98 \\\\ = \frac{1}{\cancel{2}}\times\frac{\cancel{2}}{10} \times 98 \\\\ = \frac{98}{10} = 9.8~Joule (J)$