Question

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

Answer 1

Answer:

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:

        $\Delta K + \Delta U = 0$

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

       $\Delta K = K_{f} - K_{0} = K_{f} - 0 = K_{f} = \frac{1}{2} * m* v_{f} ^{2} (1)$

• 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:

       $\Delta U = U_{f} - U_{0} = 0 - (m*g*h) = -m*g*h (2)$

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

        $v =\sqrt{2*g*h} = \sqrt{2*9.8 m/s2*23.0m} = 21.2 m/s$

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