Question

A ball of mass 200 g falls to the ground from a height of 2.5 m and rises to a height of 0.4 m on
hitting the ground. If the time of contact of the ball with the ground is 0.02 s. What is the force exerted by ground on ball​

Answer 1

force exerted by ground =ma

=0.2×10

=2N

Answer 2

The force exerted by ground on ball is 98 N.

Explanation:

Given that,

Mass of the ball, m = 200 g = 0.2 kg

It  falls to the ground from a height of 2.5 m and rises to a height of 0.4 m on  hitting the ground.

Using conservation of energy to find initial and final velocity.

Initial velocity,

$u=\sqrt{2gh} \\\\u=\sqrt{2\times 9.8\times 2.5} \\\\u=7\ m/s$

Final velocity,

$v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 0.4} \\\\v=2.8\ m/s$

Final speed, v = -2.8 m/

Time of contact, t = 0.02 s

Let F is the force exerted by ground on ball​. It is given given using the concept of impulse as :

$F\times t=m(v-u)\\\\F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{0.2\times (-2.8-7)}{0.02}\\\\F=-98\ N$

So, the force exerted by ground on ball is 98 N.

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