Solveeit Logo

Question

Physics Question on projectile motion

The angle for which maximum height and horizontal range are same for a projectile is

A

3232^\circ

B

4848^\circ

C

7676^\circ

D

8484^\circ

Answer

7676^\circ

Explanation

Solution

Given H = R
u2sin2θ2g=u2sin2θg\therefore \frac{ u^2 \, \sin^2 \, \theta }{ 2 g } = \frac{ u^2 \, \sin \, 2 \, \theta }{ g }
sin2θ2=2sinθcosθ\frac{ \sin^2 \, \theta }{ 2 } = 2 \, \sin \, \theta \, \cos \, \theta
sinθcosθ=4anθ=4\frac{\sin \, \theta }{ \cos \, \theta } = 4 \Rightarrow an \, \theta = 4
θ=tan1(4),θ76\therefore \theta = \tan^{ - 1} \, ( 4 ), \, \theta \approx 76^\circ