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 st

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

in progress 0
Nguyệt Ánh 5 years 2021-09-05T10:50:01+00:00 1 Answers 1448 views 0

Answers ( )

  1. Verity
    0
    2021-09-05T10:51:06+00:00
    Reply

    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.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )