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Question

# The trajectory of the projectile near the surface of the earth is given as y=2x−9x2. If it was launched at an angle θ0 with speed v0 then (g=10 m/s2)

A
θ0=sin115 and v0=35 ms1
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B
θ0=cos125 and v0=35 ms1
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C
θ0=cos115 and v0=53 ms1
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D
θ0=sin125 and v0=35 ms1
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Solution

## The correct option is C θ0=cos−11√5 and v0=53 ms−1Given, y=2x−9x2 On comparing with, y=xtanθ−gx22v20cos2θ We have, tanθ=2 or cosθ=1√5 [∵cosθ=1√1+tan2θ] And, g2v20cos2θ=9 ⇒102v20(1/√5)2=9 ∴v0=53 m/s Hence, (C) is the correct answer.

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