Question

A particle is projected up with a speed of 25 m/s from the ground. Find the
distance covered by the stone during 3rd second of its motion?

Answer 1

Answer:

Given

u=25m/s

t=3 sec

a(due to gravity) is approximately -10m/s^2

v=?

s=?

So, using the first equation of motion we get,

v=u + at

= 25 + -10 x 3

= 25 - 30

=-5m/s

Now placing this value in the third equation of motion we get,

2as=v^2-u^2

2 x -10 x s = (-5)^2 - (25)^2

-20s= 25-625

-20s = -600

s= -600/-20=30 m

Therefore the distance covered by the stone in the 3rd second of motion is 30 m

Hope this will help you.

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