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A nut fell off a rocket launched vertically upwards with a uniform speed at a height of 50m from ground level. If it takes 6 seconds for the nut to hit the ground, what was the velocity of the rocket just before the nut fell off the rocket? A) 40 m/s B) 20 m/s C) 10 m/s D) 5 m/s

Answer :

Final answer:

The velocity of the rocket just before the nut fell off is 58.8 m/s.

Explanation:

To calculate the velocity of the rocket just before the nut fell off, we can use the concept of free fall. The nut falls for 6 seconds, which means it was in free fall for 6 seconds. During free fall, the acceleration due to gravity is constant at 9.8 m/s². So, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

In this case, the initial velocity of the nut is the same as the velocity of the rocket just before the nut fell off. The acceleration is -9.8 m/s² (taking into account the direction). The time is 6 seconds. Plugging in these values, we have v = u + (-9.8)(6). Solving for u, we get u = v + 9.8(6).

Since the nut falls from a height of 50m, it reaches a final velocity of 0 m/s at the ground level. Therefore, the final velocity of the rocket just before the nut falls off is 9.8(6) = 58.8 m/s. So, the answer is not among the given choices.

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