JEE MAIN - Physics (2023 - 8th April Morning Shift - No. 12)
Two projectiles A and B are thrown with initial velocities of $$40 \mathrm{~m} / \mathrm{s}$$ and $$60 \mathrm{~m} / \mathrm{s}$$ at angles $$30^{\circ}$$ and $$60^{\circ}$$ with the horizontal respectively. The ratio of their ranges respectively is $$\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)$$
$$4: 9$$
$$2: \sqrt{3}$$
$$\sqrt{3}: 2$$
$$1: 1$$
Explanation
The range of a projectile launched with an initial velocity $v$ at an angle $\theta$ with respect to the horizontal is given by:
$R = \frac{v^2 \sin(2\theta)}{g}$,
where $g$ is the acceleration due to gravity.
Let's calculate the ranges of projectiles A and B:
For projectile A, $v = 40 \, \text{m/s}$ and $\theta = 30^\circ$, so:
$R_A = \frac{(40)^2 \sin(2 \times 30)}{10} = 4 \times 40 = 160 \, \text{m}$.
For projectile B, $v = 60 \, \text{m/s}$ and $\theta = 60^\circ$, so:
$R_B = \frac{(60)^2 \sin(2 \times 60)}{10} = 6 \times 60 = 360 \, \text{m}$.
Therefore, the ratio of their ranges is $R_A : R_B = 160 : 360 = 4 : 9$.
Comments (0)
