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Question

The time of flight of projectile is 10 second and its range is 500m. The maximum height reached by will be (g=10m/s2)

A
25m
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B
50m
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C
82m
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D
125m
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Solution

The correct option is D 125m
Given,
T=10sec
R=500m
g=10m/s2
Maximum height, H=u2sin2θ2g.. . . . . . . . . . . . . .(1)

Time of flight, T=2usinθg
10=2usinθ10

usinθ=50

From equation (1),
H=50×502×10

H=125m
The correct option is D.

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