Physics
Kinematics
Projectile Motion

Kinematics · JEE & NEET Physics

Projectile Motion: notes and previous year questions

Anything thrown and left to gravity: horizontal and oblique throws, time of flight, height and range, the trajectory, and throws from a height or up a slope.

48 JEE Main questions (2003–2026)
8 NEET questions (2002–2023)

Projectile Motion in short

A projectile moves under gravity alone: its acceleration is g downward and its path is a parabola.
The horizontal velocity stays constant; the vertical velocity changes by g every second.
Horizontal and vertical motions are independent and share only time.
A ball thrown sideways lands at the same time as one dropped from the same height.

1What projectile motion is

A projectile is any object that is thrown and then moves under gravity alone (air resistance ignored). Its acceleration is always $g$ , straight down; we take $g = 10\ \text{m/s}^2$ .

Horizontal motion: no acceleration, so the horizontal velocity stays constant.
Vertical motion: constant acceleration $g$ downward.
Path: a parabola.

Without gravity, the ball would fly along the straight line of the throw. Gravity pulls it below that line by $\tfrac{1}{2}gt^2$ , bending the path into a parabola.

2Horizontal projection

Thrown horizontally at speed $u$ from height $h$ : $u_x = u$ and $u_y = 0$ . Sideways, $x = ut$ ; downward, $y = \tfrac{1}{2}gt^2$ and $v_y = gt$ .

T = \sqrt{\frac{2h}{g}}

The time of flight depends only on the height, not on u.

R = uT = u\sqrt{\frac{2h}{g}}

v_{\text{final}} = \sqrt{u^2 + 2gh}

A ball dropped and a ball thrown sideways at the same moment from the same height land together.

3Oblique projection

Thrown at speed $u$ and angle $\theta$ above the ground, split $u$ into components:

u_x = u\cos\theta, \quad u_y = u\sin\theta

Then at any time $t$ :

$x = (u\cos\theta)\,t$ and $v_x = u\cos\theta$ (constant);
$y = (u\sin\theta)\,t - \tfrac{1}{2}gt^2$ and $v_y = u\sin\theta - gt$ .

Example: with $u_x = 16$ m/s and $u_y = 12$ m/s ( $u = 20$ m/s, $\theta \approx 37°$ ), after 2 s $v_y = 12 - 20 = -8$ m/s: 8 m/s downward.

4Time of flight, maximum height and range

At the top $v_y = 0$ , so $u\sin\theta - gt = 0$ : the time to the top is $u\sin\theta/g$ . Coming down takes as long as going up.

T = \frac{2u\sin\theta}{g}

For the height, use $v_y^2 = u_y^2 - 2gH$ with $v_y = 0$ :

H_{\max} = \frac{u^2\sin^2\theta}{2g}

The range is the horizontal speed times the time of flight: $u\cos\theta \times 2u\sin\theta/g$ . Since $2\sin\theta\cos\theta = \sin 2\theta$ :

R = \frac{u^2\sin 2\theta}{g}

5Maximum range and complementary angles

$R = u^2\sin 2\theta/g$ is largest when $\sin 2\theta = 1$ , that is at $\theta = 45°$ :

R_{\max} = \frac{u^2}{g}

At 45° the height is R_max / 4. At 20 m/s: R_max = 40 m.

Angles $\theta$ and $90° - \theta$ give the same range, because $\sin 2\theta = \sin(180° - 2\theta)$ . The steeper throw goes higher and stays up longer:

\frac{H_1}{H_2} = \tan^2\theta, \quad T_1T_2 = \frac{2R}{g}

Angle	Range	Height	Time
30°	$R$	$H_1$	$T_1$
60°	$R$ (the same)	$H_2 > H_1$	$T_2 > T_1$

6The trajectory and the velocity

Remove time: $t = x/(u\cos\theta)$ from the $x$ equation, put into the $y$ equation:

y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}

Of the form y = ax − bx²: a parabola. The first term is the line of the throw; the second is how far gravity pulls the ball below it.

y = x\tan\theta\left(1 - \frac{x}{R}\right)

y = 0 at x = 0 and at x = R.

The speed at height $h$ and at time $t$ , and its direction $\alpha$ with the ground:

v = \sqrt{u^2 - 2gh}

v = \sqrt{u^2\cos^2\theta + (u\sin\theta - gt)^2}

\tan\alpha = \frac{u\sin\theta - gt}{u\cos\theta}

At the same height the speed is the same going up and coming down, and it is smallest at the top, where only $u\cos\theta$ is left. Thrown at 20 m/s with $u_x = 12$ and $u_y = 16$ m/s: 12 m/s at the top; at 7.2 m up, $v = \sqrt{400 - 144} = 16$ m/s.

7Special cases

From a height $h$ . At landing $y = -h$ : solve $-h = (u\sin\theta)T - \tfrac{1}{2}gT^2$ and keep the positive root:

T = \frac{u\sin\theta + \sqrt{u^2\sin^2\theta + 2gh}}{g}

From a 40 m tower at 20 m/s, 30° up: T = (10 + √900)/10 = 4 s, range ≈ 17.3 × 4 ≈ 69 m.

Up an inclined plane of angle $\alpha$ , thrown at $\theta$ from the horizontal (so at $\theta - \alpha$ above the slope):

R = \frac{2u^2\sin(\theta - \alpha)\cos\theta}{g\cos^2\alpha}

\theta_{\text{best}} = 45° + \frac{\alpha}{2}

For a 30° slope: 60°. For a 20° slope: 55°.

R_{\max} = \frac{u^2}{g(1 + \sin\alpha)}

Summary

Key ideas

A projectile moves under gravity alone: its acceleration is g downward and its path is a parabola.
The horizontal velocity stays constant; the vertical velocity changes by g every second.
Horizontal and vertical motions are independent and share only time.
A ball thrown sideways lands at the same time as one dropped from the same height.
At the top only the vertical velocity is zero; the speed there is u cos θ and the acceleration is still g.
Time of flight, maximum height and range come from the vertical and horizontal equations separately.
The range is largest at 45°, and complementary angles give the same range.
The steeper of two complementary throws goes higher and stays up longer.
At the same height the speed is the same going up and coming down.
From a height, solve the quadratic for T and keep the positive root; up a slope, the best angle is 45° + α/2.

Every equation

Components: $u_x = u\cos\theta, \; u_y = u\sin\theta$
Horizontal: time of flight: $T = \sqrt{2h/g}$
Horizontal: range: $R = u\sqrt{2h/g}$
Horizontal: landing speed: $v = \sqrt{u^2 + 2gh}$
Oblique: time of flight: $T = \frac{2u\sin\theta}{g}$
Oblique: maximum height: $H_{\max} = \frac{u^2\sin^2\theta}{2g}$
Oblique: range: $R = \frac{u^2\sin 2\theta}{g}$
Maximum range: $R_{\max} = u^2/g \text{ at } 45°$
Complementary angles: $H_1/H_2 = \tan^2\theta$
Complementary angles: $T_1T_2 = 2R/g$
Trajectory: $y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$
Trajectory with R: $y = x\tan\theta\,(1 - x/R)$
Speed at height h: $v = \sqrt{u^2 - 2gh}$
Direction of velocity: $\tan\alpha = \frac{u\sin\theta - gt}{u\cos\theta}$
From a height: $T = \frac{u\sin\theta + \sqrt{u^2\sin^2\theta + 2gh}}{g}$
Up an incline: range: $R = \frac{2u^2\sin(\theta - \alpha)\cos\theta}{g\cos^2\alpha}$
Up an incline: best angle: $\theta = 45° + \alpha/2$

Previous year questions with solutions

Real JEE and NEET questions on projectile motion. Try each one before you open the solution.

Q1JEE Main 2026One correct option

The two projectiles are projected with the same initial velocities at the ${15}^{\circ }$ and ${30}^{\circ }$ with respect to the horizontal. The ratio of their ranges is $1:x$ . The value of $x$ is

A $\sqrt{2}$
B $\sqrt{3}$
C $2\sqrt{3}$
D $\frac{1}{\sqrt{2}}$

Show answer and solution

Answer: Option B

$15^{\circ}$ and $30^{\circ}$ add to $45^{\circ}$ , not $90^{\circ}$ , so this is not a complementary pair and the ranges do not tie. With the speed shared, $R \propto \sin 2\theta$ , so the ratio is $\sin 30^{\circ} : \sin 60^{\circ} = \tfrac{1}{2} : \tfrac{\sqrt{3}}{2} = 1 : \sqrt{3}$ , giving $x = \sqrt{3}$ . Double the angle before taking the sine — comparing $\sin 15^{\circ}$ with $\sin 30^{\circ}$ is the slip this one is set to catch.

Q2NEET 2023One correct option

A bullet is fired from a gun at the speed of $280{\mathrm{ms}}^{-1}$ in the direction ${30}^{\circ }$ above the horizontal. The maximum height attained by the bullet is

$(g=9.8{\mathrm{ms}}^{-2},\sin {30}^{\circ }=0.5)$ :-

A2000 m
B1000 m
C3000 m
D2800 m

Show answer and solution

Answer: Option B

The question hands you $g = 9.8\ \mathrm{m/s^{2}}$ , so use that rather than the usual $10$ .

The upward component is $280 \times 0.5 = 140\ \mathrm{m/s}$ , and $H = \dfrac{140^{2}}{2 \times 9.8} = \dfrac{19600}{19.6} = 1000\ \mathrm{m}$ .

The numbers are chosen to divide cleanly, which is the sign that you have taken the right component: had you used the whole $280\ \mathrm{m/s}$ you would have got $4000\ \mathrm{m}$ , which is not on offer.

Q3JEE Main 2026One correct option

If $x$ and $y$ coordinates of a projectile as a function of time $(t)$ are given as $24t$ and $43.6t-4.9t^{2}$ , respectively, then the angle (in degrees) made by the projectile with horizontal when $t=2 s$ is $\_\_\_\_$ .

Show answer and solution

Answer: Option B

Differentiate each coordinate to get its velocity component. From $x = 24t$ , $v_x = 24\ \mathrm{m/s}$ and it stays there. From $y = 43.6t - 4.9t^{2}$ , $v_y = 43.6 - 9.8t$ , which at $t = 2\ \mathrm{s}$ is $43.6 - 19.6 = 24\ \mathrm{m/s}$ . The components are equal, so $\tan\alpha = \tfrac{24}{24} = 1$ and $\alpha = 45^{\circ}$ . The $4.9$ in front of $t^{2}$ is $\tfrac{1}{2}g$ with $g = 9.8\ \mathrm{m/s^{2}}$ , which is the value this question is built on.

Practice questions, easy to hard

Three questions from the projectile motion practice ladder: one easy, one medium, one hard.

Q4One or more correct options

A ball is launched from level ground at speed $u$ and angle $\theta$ , and lands back on the ground.

Select every statement that is true of its flight.

AThe horizontal component of its velocity is the same at landing as at launch
BGravity changes the vertical component only
CThe time spent rising equals the time spent falling
DThe time of flight grows if the horizontal component is made larger

Show answer and solution

Answer: Options A, B, C

Nothing pushes or pulls sideways, so $u\cos\theta$ survives the whole flight — that is A, and B is the same fact said from the other end. The vertical motion is a straight-up throw, whose rise and fall are mirror images, which gives C. D is the standing trap of this rung: $T = \dfrac{2u\sin\theta}{g}$ never mentions the horizontal part.

Q5One or more correct options

A projectile passes the same height twice, once on the way up and once on the way down, and the two halves are mirror images. If it is at that height at $t_1$ and again at $t_2$ , the top sits midway between them and $t_1 + t_2 = T$ , the whole time of flight.

A ball thrown from the ground is at the same height at $t = 2\ \mathrm{s}$ and at $t = 6\ \mathrm{s}$ . Select every statement that follows.

AIts time of flight is $8\ \mathrm{s}$
BIt is at the top of its path at $t = 4\ \mathrm{s}$
CIts vertical component at $6\ \mathrm{s}$ is the same size as at $2\ \mathrm{s}$ , pointing the other way
DIts horizontal component at $6\ \mathrm{s}$ is smaller than at $2\ \mathrm{s}$

Show answer and solution

Answer: Options A, B, C

$T = t_1 + t_2 = 2 + 6 = 8\ \mathrm{s}$ , and the top is halfway, at $4\ \mathrm{s}$ . That alone fixes the launch: the vertical component has to die away in those $4\ \mathrm{s}$ , so $u\sin\theta = 10 \times 4 = 40\ \mathrm{m/s}$ . Then $v_y = 40 - 20 = +20\ \mathrm{m/s}$ at $2\ \mathrm{s}$ and $40 - 60 = -20\ \mathrm{m/s}$ at $6\ \mathrm{s}$ : same size, opposite way, which is why the heights agree. D is the standing trap — the horizontal component is the one thing that never changes.