Physics, asked by ashok20, 1 year ago

the velocity at the maximum height of a projectile is half of its initial velocity of projection. the angle of projection is

Answers

Answered by 7509806468anusouyyiu
7
As ucosθ0=u/2ucos⁡θ0=u/2θo=60∘θo=60∘So R=u2sin(2×60∘)gR=u2sin⁡(2×60∘)g=3–√2u2g

Answered by JunaidMirza
30
At maximum height velocty = uCosθ

uCosθ = u/2
Cosθ = 1/2
θ = 60°

Angle of projection is 60°
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