Question

Q.) what is the angle of projection for a projectile motion whose range R is n times the maximum height H ?

Answer 1

R = nH

2u²CosθSinθ/g = nu²Sin²θ/(2g)

4Cosθ = nSinθ

Tanθ = 4/n

θ = Tan⁻¹(4/n)

Angle of projection is Tan⁻¹(4/n)

Answer 2

ANGLE IS tan^(-1)[4/n]

Range is n times the max height .

So, we can say:

$R = n \times H$

• Velocity of projection be u, angle of projection is $\theta$ , gravity is g.

$\implies \dfrac{ {u}^{2} \sin(2 \theta) }{g} = n \times \dfrac{ {u}^{2} { \sin}^{2}( \theta) }{2g}$

$\implies \dfrac{ {u}^{2} 2\sin( \theta) \cos( \theta) }{g} = n \times \dfrac{ {u}^{2} { \sin}^{2}( \theta) }{2g}$

$\implies \dfrac{ 2\sin( \theta) \cos( \theta) }{g} = n \times \dfrac{ { \sin}^{2}( \theta) }{2g}$

$\implies \: \tan( \theta) = \dfrac{4}{n}$

$\implies \: \theta = { \tan}^{ - 1} ( \dfrac{4}{n} )$

So, the angle is tan^(-)[4/n]