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Question

# Two projectiles are projected at angles (θ) and (π2−θ) to the horizontal respectively with same speed 20m/s. One of them rises 10 m higher than the other. Find the angle of projection θ.

A

30
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B

45
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C

37
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D

53
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Solution

## The correct option is A 30∘Maximum height for a projectile is given by the formula H=u2sin2θ2g Let, h1 be the maximum height for 1st projectile. So, h1=u2sin2θ2g Given, u=20 m/s; g=10 m/s2 ⇒h1=202×sin2θ2×10 ⇒h1=20sin2θ Let h2 be the maximum height for second projectile. So, h2=u2sin2(π2−θ)2g Substituting the values, h2=202×sin2(π2−θ)2×10 ⇒h2=20cos2θ Given that, h2−h1=10 m ⇒20cos2θ−20sin2θ=10 ⇒cos2θ−sin2θ=1/2 ⇒cos2θ=12 ⇒2θ=cos−112 ⇒2θ=60∘ ⇒θ=30∘ Hence, option (a) is the correct choice.

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