Question

37. If a body of mass 3 kg is dropped from the top of
a tower of height 250 m, then its kinetic energy
after 3 s will be
(1) 1126 J
(2) 1048 J
(3) 735 J
(4) 1296.5 J
​

Answer 1

Given:-

→Mass of the body = 3kg

→Height of the tower = 250m

→Time = 3s

To find:-

→Kinetic energy of the body after 3s

Solution:-

In this case:-

•Acceleration due to gravity of the body will be +9.8m/s² as the body is freely falling.

•Initial velocity of the body will be zero as the body was initially at rest.

Firstly, let's calculate the velocity of the body after 3 seconds.

By using the 1st equation of motion, we get:-

=> v = u+at

=> v = 0+(9.8)3

=> v = 29.4m/s

Thus, velocity of the body after 3s is 29.4m/s.

Now, we know that:-

=> K.E. = 1/2mv²

=> K.E. = 1/2×3×(29.4)²

=> K.E. = 1.5×864.36

=> K.E. = 1296.5 J

Hence,kinetic energy of the body after 3s will be 1296.5 J (Option.4)

Answer 2

Answer:

1296.5 J

Explanation:

mass=3kg

height=250

time=3s