Question

If a body of mass 3 kg is dropped from top of a tower of height 25m then the kinetoc energy after 3 sec is

Answer 1

IIn this question, you need to verify if the body will still be mid air or reach the ground 3 seconds after the drop.
Using formula s = ut + 1/2 gt^2,
250 = 0*t + 1/2 * 9.8 * t^2
t = 7.14 secs
So the body will reach the ground 7.14 sec after when dropped from height of 250 m. So at 3 sec, it will still be in air.
Velocity after 3 seconds = u + gt = 0 + 9.8*3 = 29.4 m/s
So, K.E. after 3 sec will be: 1/2 * m * v^2
= 1/2 * 3 * (29.4)^2 = 1296.54 J