Question 41: 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: