Projectile MotionJEE Main

Concept

A projectile moves under gravity alone after launch. Its horizontal and vertical motions are independent: horizontal velocity stays constant (no horizontal force), while vertical motion is uniformly accelerated by $g$. The path traced is a parabola.

Key formula

Derivation

Resolve the launch velocity: $u_x = u\cos\theta$, $u_y = u\sin\theta$.

Vertical: $y = u_y t - \tfrac{1}{2}gt^2$. Setting $y=0$ for level ground gives time of flight $T = \dfrac{2u\sin\theta}{g}$.

Horizontal: $x = u_x t$. Substituting $T$: $R = u\cos\theta \cdot \dfrac{2u\sin\theta}{g} = \dfrac{u^2\sin 2\theta}{g}$.

At the apex $v_y = 0$, so $H = \dfrac{u_y^2}{2g} = \dfrac{u^2\sin^2\theta}{2g}$.

Scenarios to explore

• Maximum Range Launch — Classic 45° projectile maximising horizontal range.
• Cliff Edge Throw — Projectile launched from a height — range is no longer maximal at 45°.
• Lunar Projectile — Same launch, Moon gravity (1.62 m/s²) — watch the range stretch.

Real-world applications

• Sports: optimal angle for a long throw or kick.
• Ballistics and water-fountain design.
• Estimating safe stopping distances of launched objects.

JEE exam tips

• For launch from a height, use $T = \dfrac{u\sin\theta + \sqrt{u^2\sin^2\theta + 2gh}}{g}$.
• Range is symmetric: angles $\theta$ and $(90^\circ-\theta)$ give the same range on level ground.

Common mistakes

• Forgetting that horizontal velocity is constant (treating it like vertical).
• Using $\sin\theta$ instead of $\sin 2\theta$ in the range formula.
• Assuming maximum range is always at 45° — it is not when launched from a height.

Exam traps to avoid

• Two projectiles with complementary angles share a range but differ in height & time.
• Air resistance is ignored in JEE projectile problems unless explicitly stated.