Question

Ravi dropped a stone and a rubber ball from his terrace,100m high from the ground. (Neglect air resistance )
i) Will both fall with same or different acceleration ? What is the value of their accelerations ? ii) Write the formula of their acceleration .
iii) Calculate the final velocity of the rubber ball just before it reaches the ground. Will it be same in case of stone ?​

Answer 1

Answer:

As an object falls, its speed increases because it's being pulled on by gravity. The acceleration of gravity near the earth is g = -9.81 m/s^2. To find out something's speed (or velocity) after a certain amount of time, you just multiply the acceleration of gravity by the amount of time since it was let go of.

Answer 2

Given:

Ravi dropped a stone and a rubber ball from his terrace,100m high from the ground. (Neglect air resistance).

To find:

i) Will both fall with same or different acceleration ? What is the value of their accelerations ? ii) Write the formula of their acceleration .

iii) Calculate the final velocity of the rubber ball just before it reaches the ground. Will it be same in case of stone ?

Calculation:

Question i)

Yes, they will fall with same acceleration towards earth surface , if air resistance is neglected.

• Value of Gravitational Acceleration for both the objects will be approximately 9.8 m/s².

The general expression for the gravitational acceleration acting on the bodies will be:

$\boxed{ \bold{ \therefore \: g = \dfrac{GM}{{R}^{2} } \approx \: 9.8 \: m {s}^{ - 2} }}$

• G = Universal Gravitational Constant, M = mass of earth, R = radius of earth.

Question ii)

Let the final velocity for the objects be v:

$\therefore \: {v}^{2} = {u}^{2} + 2gh$

$\implies \: {v}^{2} = {(0)}^{2} + 2(9.8)(100)$

$\implies \: {v}^{2} = 1960$

$\implies \: v = \sqrt{1960}$

$\implies \: v = 44.27 \: m {s}^{ - 1}$

So, final velocity for both objects (both stone and rubber ball) will be 44.27 m/s.