Question

A stone with weight w is thrown vertically upwards into air from ground level with initial velocity v0. If a constant force f due to air drag acts on the stone throughout ight, the speed of the stone just before impact with the ground is

Answer 1

Answer:

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Explanation:

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Answer 2

speed of the stone just before impact with ground is v0/√(1 + f/w).

weight of stone = w

initial velocity of stone = v0

a constant force f due to air drag acts on the stone .

it can be solved easily with help of law of conservation of energy

initial energy = kinetic energy of stone = 1/2 mv0² ......(1)

final energy = potential energy + energy due to air drag

let stone reaches h height from the ground.

so, final energy = (f + w)h .....(2)

initial energy = final energy

1/2 mv0² = (f + w)h

⇒h = mv0²/2(f + w) = mv0²/2w(1 + f/w)

we know, w = mg

so, h = v0²/2g(1 + f/w)

now, speed of stone just before the impact with the ground , v = √(2gh)

= √{2 × g × v0²/2g(1 + f/w)}

= v0/√(1 + f/w)

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