Question

the horizontal range of projectile is 4 root 3 times of its maximum height. the angle of projection will be


​

Answer 1

Answer:

The correct option is D

30

Answer 2

$\text{Let the range of the horizontal projectile be }R$

$\text{Let the max height attained by it be }H_{max}$

$\text{It is given that }R = 4\sqrt3\times H_{max} \text{ ------ (i)}$

$\text{We know that range}(R) = \dfrac{u^{2}sin2\theta}{g} \text{ ------ (ii)}$

$\text{We also know that max height}(H_{max}) = \dfrac{u^{2}sin^{2} \theta}{2g} \text{ ------ (iii)}$

$\implies \text{From (i). (ii) and (iii)}$

$\implies \dfrac{u^{2}sin2\theta}{g} = 4\sqrt3 \times \dfrac{u^{2}sin^{2}\theta}{2g}$

$\implies sin2\theta = 2\sqrt3 \times sin^{2}\theta$

$\text{We know that }sin2\theta = 2sin\theta cos\theta$

$\implies 2sin\theta cos\theta = 2\sqrt3 \times sin^{2}\theta$

$\implies cos\theta = \sqrt3 \times sin\theta$

$\implies cot\theta = \sqrt3$

$\implies \boxed{\theta = \dfrac{\pi}{6} = 30^{\circ}}$