A rocket moving up at 210 m/s drops a segment with mass 3 kg. If the height of the rocket was 7600 m what is the kinetic energy of the segment when it reached the ground. (Assume constant g = 10 m/s^2)
1 Answer | Add Yours
The rocket is moving upwards at 210 m/s and at a height of 7600 m a segment with mass 3 kg falls from it.
As the rocket is moving upwards and the gravitational acceleration is in the downward direction, the initial velocity of the segment is -210 m/s. The velocity of the segment when it impacts the ground is given by v^2 - u^2 = 2*a*s, where u = -210, a = 10 and s = 7100.
v^2 = 210^2 + 20*7100 = 186100
The kinetic energy of a body with mass m moving at a velocity v is (1/2)*m*v^2. Using this, the kinetic energy of the segment when it strikes the ground is equal to 279150 J.
We’ve answered 300,908 questions. We can answer yours, too.Ask a question