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Question

The distance travelled by a body during last second of its upward journey is d, when the body is projected with certain velocity vertically up. If the velocity of projection is doubled, the distance travelled by the body during the last second of its upward journey is

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Solution

The correct option is **D** d

Let the speed of projection be u and the time taken to reach maximum height be t.

St=ut−0.5gt2

St−1=u(t−1)−0.5g(t−1)2

Let the speed of projection be u and the time taken to reach maximum height be t.

St=ut−0.5gt2

St−1=u(t−1)−0.5g(t−1)2

Solving these two equations, we get

St−St−1=u−gt+g2

St−St−1=u−gt+g2

which is the distance traveled in the last second.

For a vertically projected body, u=gt (because 't' is the time when it reachs max. height), hence

st−st−1=g2

st−st−1=g2

This shows that the distance traveled by a body projected vertically up is independent of the initial speed.

So, even if the body is projected with double the speed, the body covers the same distance, d in the last second.

So, even if the body is projected with double the speed, the body covers the same distance, d in the last second.

Hence, option D is correct.

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