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Question

When a projectile is thrown at angle θ with the vertical it reaches the maximum height D. The time of flight of the projectile is

A
Dg
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B
2Dg
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C
4Dg
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D
8Dg
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Solution

The correct option is D 8Dg
Lets consider, u= initial velocity of the projectile
θ= angle of projection
Maximum height, D=u2sin2θ2g
u2sin2θ=2gD
usinθ=2gD. . . . . . .(1)
Time of flight of the projectile,
T=2usinθg
T=22gDg
T=8gDg2
T=8Dg
The correct option is D.

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