Question

Horizontal Range of Projectile is twice of the max. height, then what is the angle of projection?

Answer 1

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Here is your answer


By formula, of projectile motion


Range = U ^2 Sin theta /g

max height = u ^2 sin ^2 theta / 2g


Here U is the initial velovity of projectile , theta is angle of projectile with horizontal, g is acceleration due to gravity

now , according to question

R= 2H

U^2 sin ^2 = 2u^2 sin^2 theta /2g

SIN 2 THETA = sin ^2 theta

2 Sin theta × cos theta = sin^2 theta

2cos theta = sin theta

tan theta =2


sin theta =2 / underrot 5

cos theta = 1 /underrot 5


Now ,

Range = U^2 sin 2 theta / g

u^2 ( 2 sin theta ×cos theta ) /g

U^2 ( 2×2 / underroot 5 × 1 /underroot 5 ) /g

= 4 u^2 /5 g


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