Question

the angle of projection of projectile is 30 what is the ratio of horizontal range and maximum height

Answer 1

R=4 H cot theta. R/H= 4cot theta. theta = 30. so ans is 4 root3

Answer 2

Answer:

$\dfrac{R}{H}=4\sqrt3$

Explanation:

It is given that,

the angle of projection is, $\theta=30^{\circ}$

The horizontal range of a projectile is given by :

$R=\dfrac{v^2sin2\theta}{g}$..........(1)

The maximum height attained by the projectile is given by :

$H=\dfrac{v^2sin^2\theta}{2g}$..............(2)

On dividing equation (1) and (2) as :

$\dfrac{R}{H}=\dfrac{\dfrac{v^2sin2\theta}{g}}{\dfrac{v^2sin^2\theta}{2g}}$

$\dfrac{R}{H}=4\times cot\theta$

$\dfrac{R}{H}=4\times cot(30)$

$\dfrac{R}{H}=4\sqrt3$

Hence, this is the required solution.