The range of a projectile is given by $R = \frac{u^2 \sin(2\theta)}{g}$. This is derived by finding the time of flight $T = \frac{2u\sin\theta}{g}$ and horizontal distance $R = u\cos\theta \cdot T$.
Back to JEE Main 2025 Mechanics
JEE Main 2025 — Physics Mechanics
medium
mcq
2025
Official previous-year question
Verified 30 May 2026.
Question
The range of a projectile launched with initial velocity $u$ at angle $\theta$ with the horizontal is given by:
Options
- A
$R = \frac{u^2 \sin(2\theta)}{g}$
- B
$R = \frac{u^2 \sin(\theta)}{g}$
- C
$R = \frac{u^2 \cos(2\theta)}{g}$
- D
$R = \frac{2u^2 \sin(\theta)}{g}$
Solution
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
JEE Main Mechanics in other years
More JEE Main Mechanics Questions
A particle of mass 2 kg is projected vertically upward with a speed of 30 m/s. The maximum height reached by the particle is (g = 10 m/s²):
2026
easy
mcq
A ball is projected at 45° with speed 20 m/s. Find the maximum height reached (in metres). Take g = 10 m/s².
2024
easy
numerical_value
A particle moves in a circle of radius R with constant speed v. Its centripetal acceleration is
2024
easy
mcq
A particle moves in a circle of radius R with constant angular velocity ω. If the centripetal acceleration is 4 m/s² when R = 2 m, what is the angular velocity?
2024
easy
mcq
The dimension of Planck constant is same as
2024
medium
mcq