Question

a ball of mass m is dropped from height h.average power for complete motion will be?pls answer​

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

The average power for complete motion will be $\frac{mg}{2} \sqrt{gh}$.

Explanation:

Given,

A ball of mass 'm' is dropped from height 'h'.

To find,

The average power of complete motion(P).

Calculation,

We know that power is defined as the energy per unit of time.

P = E/t

First, we calculate the energy of the ball before it hits the ground:

i.e. E = potential energy decreased by the ball.

E = mgh

Then we calculate the time taken(t) by the ball to reach the ground.

s = ut + 1/2at²

Here s = h, u = 0, a = g, t = t

h = 1/2(gt²)

$t = \sqrt{\frac{2h}{g} }$

Hence,

$P = \frac{mgh}{\sqrt{\frac{2h}{g} } }$

P  = mg/2 × (√(gh))

Therefore, the average power for complete motion will be $\frac{mg}{2} \sqrt{gh}$.

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