A 4.80-kg watermelon is dropped from rest from the roof of a 23.0-m -tall building and feels no appreciable air resistance Just before it strikes the ground, what is the watermelon's speed

v = 21.2 m/s

Explanation:

We can find the watermelon's speed just before it strikes the ground, in different ways.One option is just apply the conservation of energy principle; assuming no appreciable air resistance, total mechanical energy must remain the same.So, we can write the following equation:

       

If the watermelon started from rest at the roof of the building, the change in kinetic energy must be as follows:

       

This value must be equal in magnitude, to the change in the gravitational potential energy.If we take as zero reference level the ground level, we can find the change in gravitational potential energy as follows:

       

From (1) and (2) we can solve for v, as follows:

       

So, the watermelon's speed just before it strikes the ground, is 21.2 m/s.

The best description is: . nobody has any idea what q is.

Answermuscle strength;