Question

Two projectiles are projected with the same velocity
If one is projected at an angle of 30° and the other at
60° to the horizontal. The ratio of maximum heights
reached, is:​

Answer 1

Answer:

The ratio is 1/3

Explanation:

The initial velocity for both the projectiles is the same. Let it be u m/s.

The maximum height of a projectile is given by,

$h = \frac{{u}^{2} {sin}^{2} \gamma }{2g}$

Where, gamma is the angle of projection, u is the initial velocity and g is the acceleration due to gravity.

Considering one angle at 30° and other as 60°, the ratio will be,

$\frac{h1}{h2} = \frac{ {sin}^{2} 30}{{sin}^{2} 60} = \frac{0.25}{0.75} = \frac{1}{3}$

Hence, the ratio of the maximum heights projectiles lauched from angles 30° and 60° is 1/3.

Answer 2

Answer:

The ratio of maximum heights of projectiles reached is 1 : 3.​

Explanation:

When a body is projectile at inclination of angle θ from the ground. The maximum height (H) achieved by a projectile is given by:

$H= \frac{u^2sin^2\theta}{2g}$                                                               ...................(1)

Where u is the initial velocity of the projectile.

Now, we have given two angle, $\theta_{1}=30^o$  and $\theta_{2}=60^o$. Consider that H₁ is height of projectile due to θ₁ and H₂ is height of projectile due to angle θ₂.

The ratio of their maximum heights will be:

$\frac{H_{1}}{H_2}=\frac{u^2sin^2\theta_{1}/2g }{u^2sin^2\theta_{2}/2g}$                                       [ From the equation (1)]

$\frac{H_{1}}{H_2}=\frac{u^2sin^2\theta_{1} }{u^2sin^2\theta_{2}}$

$\frac{H_{1}}{H_2}=\frac{u^2sin^230 }{u^2sin^260}$

$\frac{H_{1}}{H_2}=\frac{sin^230 }{sin^260}$

$\frac{H_{1}}{H_2}=\frac{(\frac{1}{2})^2 }{(\sqrt{\frac{3}{2} })^2 }$

$\frac{H_{1}}{H_2}=\frac{1}{3}$

Therefore, the ratio of maximum height achieved by both projectile is 1 : 3.