Projectile is thrown with an initial velocity of v=ai^+bj^, if the range of projectile is double, the maximum height reached by it then
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Answer:
63.44°
Explanation:
Range = u²sin2∝/g
max height = u²sin²∝/2g
Range = 2 max height
u²sin2∝/g = u²sin²∝/g
sin 2∝ = sin²∝
sin 2∝ - sin²∝ = 0
2sin∝cos∝ - sin²∝ = 0
∝ ≠ 0
sin∝ = 2 cos∝
tan∝ = 2
∝ = tan∧-1 ( 2) = 63.44°
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