Question

an object is projected at the angle such that the horizondle range is 4 times the maximum height.what is the angle of projection of the object?​

Answer 1

Horizontal Range of a Projectile

$\boxed{ \bf{ R = \dfrac{ u^2 sin 2 \theta}{g}}}$

Maximum Height of a Projectile

$\boxed{ \bf{H_m = \dfrac{ {u}^{2} {sin}^{2} \theta }{2g} }}$

u → Velocity of projection

$\sf \theta$ → Angle of projection

g → Acceleration due to gravity

According to the question, Horizontal Range is 4 times the Maximum Height i.e.

$\rm \leadsto R = 4H_m \\ \\ \rm \leadsto \dfrac{ \cancel{ u^2} sin 2 \theta}{ \cancel{g}} = 4 \times \dfrac{ \cancel{{u}^{2}} {sin}^{2} \theta }{2 \cancel{g}} \\ \\ \rm \leadsto \dfrac{sin2 \theta}{ {sin}^{2} \theta} = \dfrac{4}{2} \\ \\ \rm \leadsto \dfrac{2 \cancel{sin \theta}.cos \theta}{ {sin}^{ \cancel{2}} \theta} = 2 \\ \\ \rm \leadsto \dfrac{cos \theta}{sin \theta} = \cancel{\dfrac{2}{2} } \\ \\ \rm \leadsto cot \theta = 1 \\ \\ \rm \leadsto \theta = {cot}^{ - 1} (1) \\ \\ \rm \leadsto\theta = 45 \degree$

$\therefore$ Angle of projection $\rm (\theta)$ = 45°

Answer 2

R = nH

R = 4H

n = 4

tan∅ = 4/n

tan∅ = 4/4

tan∅ = 1

∅ = 45°