Question

2. The horizontal range and the maximum height of a
projectile are equal. The angle of projection of the
projectile is

(1) = tan-1(2)
(2) = 450°
(3) = tan-1(1/4)
(4) = tan-1(4)​

Answer 1

Answer:-

$\mathsf{\theta = (4)Tan^{-1}}$

Given :-

The maximum height and the horizontal range of projectile are equal.

To find :-

The angle of the projection of the projectile.

Solution:-

The horizontal range of projectile is given by formula:-

$\huge \boxed{R = \dfrac{ u^2 Sin2\theta}{g}}$

The maximum height of projectile is given by :-

$\huge \boxed{H = \dfrac{u^2 Sin^2 \theta}{2g} }$

A/Q

→$\mathsf{R = H }$

→$\mathsf{\dfrac{u^2 Sin2\theta}{g}= \dfrac{u^2 Sin^2 \theta}{2g}}$

• Cancelling out u² and g.

→$\mathsf{Sin2 \theta = \dfrac{Sin^2 \theta }{2}}$

→$\mathsf{2Sin\theta Cos\theta = \dfrac{Sin^2 \theta }{2}}$

→$\mathsf{ 4 Sin\theta Cos \theta = Sin^2 \theta}$

• Cancelling out $Sin\theta$

→$\mathsf{4 Cos \theta = Sin \theta}$

→$\mathsf{4 =\dfrac{Sin\theta}{Cos\theta}}$

→$\mathsf{Tan\theta = 4}$

→$\mathsf{\theta = (4)Tan^{-1}}$

hence,

The angle of projection will be $\mathsf{(4)Tan^{-1}}$