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Question

For an object projected from the ground with speed u. The horizontal range is two times the maximum height attained by it. The horizontal range of the object is:

A
2u23g
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B
3u24g
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C
3u22g
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D
4u25g
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Solution

The correct option is D 4u25g
In projectile motion,
The maximum height is H=u2sin2θ2g
Horizontal range, R=u2sin2θg
According to question,
R=2H
2u2sinθ.cosθg=2u2sin2θ2g
2cosθ=sinθ
tanθ=2
We know that, 1+tan2θ=sec2θ=1cos2θ
1+22=1cos2θ
cos2θ=15
cosθ=15
And, sin2θ=1cos2θ
sin2θ=115=45
sinθ=25
Horizontal range,
R=2u2sinθ.cosθg
R=2u2g2515
R=4u25g
The correct option is D.

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