Question

what is the distance covered by a freely falling body at first three seconds of its motion g =10m/s^2​

Answer 1

Given:

• A body is falling freely.
• time = 3 seconds
• g = 10 m/s

To find:

• Distance covered in the first 3 seconds?

Calculation:

Since the question clearly mentions that the body is freely falling, we can assume that it has been dropped and its initial velocity will be zero.

Also, the ball is experiencing constant gravitational acceleration of 10 m/s².

So, applying EQUATIONS OF KINEMATICS:

$s = ut + \dfrac{1}{2} a {t}^{2}$

$\implies \: s = ut + \dfrac{1}{2} g {t}^{2}$

$\implies \: s = (0)(3) + \dfrac{1}{2} (10) {(3)}^{2}$

$\implies \: s = \dfrac{1}{2} \times 10 \times 9$

$\implies \: s = 45 \: metres$

So, distance covered in first three seconds is 45 metres.