Question

A particle released from rest from height h.Find the distance it falls through un 2 seconds,4 seconds and 6 seconds​

Answer 1

Answer:

Acceleration due to gravity (g) = \( 9.8 \frac{m}{s^2}\)

s = \(\frac{1}{2}gt^2\)

Now, at t = 1 sec

s (after 1 second) = \(\frac{1}{2}9.8 \times 1^2\)

s (after 1 second) = \( 4.9 m\)

Now, at t = 2 sec

s (after 2 second) = \(\frac{1}{2}9.8 \times 2^2\)

s (after 2 second) = \( 19.6 m\)

Now, at t = 3 sec

s (after 3 second) = \(\frac{1}{2}9.8 \times 3^2\)

s (after 3 second) = \( 44.1 m\)