2.0k views
1 vote
A bomb of mass 40 kg is dropped from an aero-plane at a height of 1 km above the ground. Whatis its K.E. (i) at the end of 10 s

User Rajul
by
8.1k points

1 Answer

3 votes

Final answer:

The kinetic energy of the bomb at the end of 10 seconds is 40590 J.

Step-by-step explanation:

To calculate the kinetic energy of an object, we can use the formula KE = 0.5 * mass * velocity^2.

Given that the mass of the bomb is 40 kg and it is dropped from a height of 1 km above the ground, we need to find its velocity at the end of 10 s.

Since the bomb is in free-fall, we can use the equation of motion s = ut + 0.5 * gt^2, where s is the height, u is the initial velocity, g is the acceleration due to gravity, and t is the time.

At the end of 10 s, the height of the bomb will be 1 km - (0.5 * 9.8 m/s^2 * 10 s)^2 = 1 km - 490 m = 510 m.

Using the equation v^2 = u^2 + 2gs, we can find the velocity at the end of 10 s as v = sqrt(2 * 9.8 m/s^2 * 510 m) = 45.05 m/s.

Now, we can calculate the kinetic energy using the formula KE = 0.5 * mass * velocity^2:

KE = 0.5 * 40 kg * (45.05 m/s)^2 = 40590 J.

User Demian Brecht
by
8.5k points