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Question

When a partical is projected from a tower of height h with horizontal with velocity u at an angle θ above the horizontal, then range of particle (on ground) is

A
[ucosθg+u2sin2θg2+2hg]ucosθ
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B
[usinθg+u2sin2θg2+2hg]ucosθ
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C
u2sin2θg2+[u2sin2θg2+2hg]cosθ
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D
u2sin2θg2
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Solution

The correct option is B [usinθg+u2sin2θg2+2hg]ucosθ

Apply kinematic equation of motion.

v2u2=2as

v=(usinθ)2+2gh

Apply kinematic equation of motion

V=U+at

t=VUa=1g((usinθ)2+2gh(usinθ))

t=(u2sin2θg2+2hg+usinθg)

Range, R=ucosθ×t=ucosθ(u2sin2θg2+2hg+usinθg)


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