QJEEM23Pha5c4: JEE Main 2023 Physics - Class 11 JEEM23Pha5c4 Solution: Two projectiles are projected at 30 ∘ and 60 ∘ with the…

Let the initial speed of both projectiles be v. The maximum height attained by a projectile can be calculated using the formula:

$$H=\frac{v^2{\sin }^2\theta }{2g}$$

where H is the maximum height, v is the initial speed, θ is the angle of projection, and g is the acceleration due to gravity.

For the projectile projected at 30°, the maximum height is:

$$H_1=\frac{v^2{\sin }^2{30}^{\circ }}{2g}=\frac{v^2\times \frac{1}{4}}{2g}=\frac{v^2}{8g}$$

For the projectile projected at 60°, the maximum height is:

$$H_2=\frac{v^2{\sin }^2{60}^{\circ }}{2g}=\frac{v^2\times \frac{3}{4}}{2g}=\frac{3v^2}{8g}$$

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 v² 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