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Question

A ball is projected with a velocity of 17 m/s at an angle θ with the horizontal. Its horizontal range is equal to its maximum height. Velocity of the particle 0.1 s after it is projected is (in m/s)
Take g=10 m/s2

A
17
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B
20
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C
10
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D
40
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Solution

The correct option is C 10
By using the relation Rtanθ=4H
Htanθ=4H (R=H)
tan θ=4
cosθ=117 and sinθ=417
Let v be the velocity of the ball after 0.1 s
vx=ux=ucosθ=17×117=1 m/s
vy=u sin θgt=17×41710×0.1=3 m/s (negative sign indicates the ball is in its downward journey)
v=v2x+v2y=12+32=10 m/s
Hence, the correct answer is option (c)

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