Question

For a projectile projected slantwise from the ground, the magnitude of its position vector with respect to the point of projection, when it is at the highest point of the path is found to be √2 two times the maximum height reached by it. Show that the angle of projection is tan⁻¹ (2).

Answer 1

Hii dear,

# Given-
|A| = √2 H

# To prove-
θ = tan⁻¹(2)

# Proof-
Consider A be the position vector for particle at Hmax.
|A| = √2 H
√(H^2 + R^2 /4) = √2 H
H^2 + R^2/4 = 2H^2
R = 2H

Putting formulas,
u^2sin2θ/g = 2×u^2(sinθ)^2 /2g
sin2θ = (sinθ)^2
2sinθ.cosθ = sinθ.sinθ
sinθ/cosθ = 2
tanθ = 2
Taking inverse,
θ = tan⁻¹(2)

Hence proved...

Hope that was useful...