Question 52: Motion in 2D & Vectors

Question

Two projectiles are projected at $30^{\circ}$ and $60^{\circ}$ with the horizontal with the same speed. The ratio of the maximum height attained by the two projectiles respectively is:

Accepted Answer

Let the initial speed of both projectiles be v. The maximum height attained by a projectile can be calculated using the formula:
$H = \frac{v^2 \sin^2 	heta}{2g}$
where H is the maximum height, v is the initial speed, ÃÂ¸ is the angle of projection, and g is the acceleration due to gravity.
For the projectile projected at 30ÃÂ°, the maximum height is:
$H_1 = \frac{v^2 \sin^2 30^{\circ}}{2g} = \frac{v^2 	imes \frac{1}{4}}{2g} = \frac{v^2}{8g}$
For the projectile projected at 60ÃÂ°, the maximum height is:
$H_2 = \frac{v^2 \sin^2 60^{\circ}}{2g} = \frac{v^2 	imes \frac{3}{4}}{2g} = \frac{3v^2}{8g}$
Now, let's find the ratio of the maximum heights:
$\frac{H_1}{H_2} = \frac{\frac{v^2}{8g}}{\frac{3v^2}{8g}} = \frac{v^2}{3v^2}$
The vÃÂ² terms cancel out, and we get:
$\frac{H_1}{H_2} = \frac{1}{3}$
Therefore, the ratio of the maximum heights attained by the two projectiles is 1 : 3

Answer

$1: \sqrt{3}$

Answer

$\sqrt{3}: 1$

Answer

1 : 3

Answer

$2: \sqrt{3}$

Motion in 2D & Vectors

Two projectiles are projected at 30^{\circ} and 60^{\circ} with the horizontal with the same speed. The ratio of the maximum height attained by the two projectiles respectively is:

Choose the correct answer: