Projectile motion

A cricketer lofts a ball over the boundary, a fountain jet arcs into the pool, a shell fired from a cannon traces a curve before it lands — every one of these is the same piece of physics wearing a different costume. Once the ball, the water droplet, or the shell leaves the launcher, exactly two things act on it in the horizontal and vertical directions: nothing, and gravity. Nothing sideways, because air resistance is ignored in this chapter; a steady $g$ downward, because that is all gravity ever does near the Earth's surface. Split the velocity into these two independent pieces and the whole curved, seemingly complicated flight collapses into two of the simplest motions in physics running side by side. This article is about projectile motion, the flagship application of two-dimensional constant-acceleration kinematics in the chapter Motion in a Plane. It builds directly on the vector addition and component-resolution tools from the previous article — here those tools finally earn their keep, because a projectile's velocity is a vector that keeps changing direction even while one of its components never changes at all. We will derive the parabolic trajectory the ball actually traces, and the three numbers every board exam wants: how long it stays in the air, how high it climbs, and how far it travels — each following from nothing more than resolving $u$ into $u\cos\theta$ and $u\sin\theta$ and applying the straight-line equations of motion you already know, once to each direction separately. By the end you should be able to look at any launch — a football free kick, a long jump, a stone thrown off a cliff — and immediately see it as two uncoupled one-dimensional problems glued together by a shared clock, $t$ .

Key takeaways

A projectile is launched, then left to gravity alone. No other force acts once it leaves the thrower's hand (air resistance neglected), so its acceleration is a constant $g$ pointing straight down.
Horizontal and vertical motions are independent. Horizontally $a_x = 0$ , so $u_x = u\cos\theta$ stays fixed for the whole flight; vertically $a_y = -g$ , so the vertical motion is exactly a ball thrown straight up and coming back down.
The path is a parabola. Eliminating time between the two component equations gives $y = x\tan\theta - \dfrac{gx^2}{2u^2\cos^2\theta}$ .
Three formulas cover level-ground launches. Time of flight $T = \dfrac{2u\sin\theta}{g}$ , maximum height $H = \dfrac{u^2\sin^2\theta}{2g}$ , and horizontal range $R = \dfrac{u^2\sin2\theta}{g}$ — all assume launch and landing are at the same height.
Range peaks at $45^\circ$ . Since $R \propto \sin2\theta$ , the range is greatest when $2\theta = 90^\circ$ ; complementary angles ( $\theta$ and $90^\circ-\theta$ ) give the same range by different paths.
At the top, velocity is purely horizontal. $v_y = 0$ there and $v_x = u\cos\theta$ throughout — the speed at the apex, $u\cos\theta$ , is the minimum speed anywhere on the path.
Key idea: resolve the launch velocity into a constant horizontal component and a uniformly decelerating-then-accelerating vertical component, and every projectile quantity — trajectory, time of flight, height, range — falls out of applying ordinary straight-line kinematics to each direction in turn.

The core insight

Projectile motion is two independent one-dimensional motions sharing one clock — constant velocity $u\cos\theta$ horizontally, and uniform acceleration $-g$ vertically — whose combination traces a parabola and fixes the time of flight, maximum height, and range.

Where it fits in the syllabus

Projectile motion is a core, exam-heavy topic of CBSE Class 11 Physics, chapter Motion in a Plane, sitting directly after vector addition/resolution and the constant-acceleration vector equations — it is the payoff application of both. The prescribed scope is precisely the trajectory equation and the $T$ , $H$ , $R$ formulas for a projectile launched at angle $\theta$ with speed $u$ under gravity alone, air resistance neglected. Relative-velocity-in-a-plane problems are touched only lightly by the board and are not part of this article; oblique projectile motion on an inclined plane and projectile motion with air drag are university-level extensions and stay firmly out of scope here.

The physical picture

Take the point of launch as the origin, $x$ horizontal and $y$ vertical with upward positive. A projectile is thrown with speed $u$ at angle $\theta$ above the horizontal, so its initial velocity has two components at the very first instant:

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

where: $u$ is the launch speed, $\theta$ is the angle the launch velocity makes with the horizontal, $u_x$ is the horizontal (constant) component, and $u_y$ is the vertical (initial) component. Physically, $u\cos\theta$ is "how much of the throw pushes it forward" and $u\sin\theta$ is "how much of the throw pushes it upward" — the same resolution-into-components idea from ordinary vector work, just applied to a velocity instead of a displacement. Gravity is the only force once the projectile leaves the hand, and gravity has no horizontal pull, so the acceleration components stay fixed for the entire flight:

a_x = 0, \qquad a_y = -g

where: $g \approx 9.8\ \text{m/s}^2$ is the acceleration due to gravity, taken downward, and the minus sign records that "up" was chosen as the positive $y$ -direction. Because $a_x = 0$ , the horizontal motion never changes speed — it is uniform velocity motion, $x = u_x t = (u\cos\theta)t$ , the simplest kinematics there is. Because $a_y = -g$ is a nonzero constant, the vertical motion is exactly the "ball thrown straight up" problem: it decelerates going up, momentarily stops at the top, then accelerates coming back down, obeying $y = u_y t - \tfrac12 g t^2$ and $v_y = u_y - gt$ . Nothing about gravity connects the two directions to each other — that decoupling is the entire trick of projectile motion, and it is why the problem is tractable with nothing more than the algebra you already know from one-dimensional motion.

Figure

A projectile launched at an angle showing the parabolic path with the initial velocity split into a constant horizontal arrow and a shrinking-then-growing vertical arrow, the apex marked with maximum height, and the base marked with time of flight and range

Eliminating time $t$ between the two component displacement equations, $x = u\cos\theta\, t$ and $y = u\sin\theta\, t - \tfrac12 g t^2$ , gives the equation of trajectory:

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

where: this is the shape of the path itself, $y$ as a function of $x$ with no time in it. Because $y$ depends on $x^2$ with a negative coefficient, the path is a downward-opening parabola — exactly the curve you see traced by any thrown object, and the algebraic proof of why every projectile "looks the same," a symmetric arc rising and falling.

Common results — trajectory to range

Derivation ATime of flight, maximum height, and range for a level-ground projectile

The goal of this derivation is to obtain the three standard results a board exam asks for — how long the projectile stays airborne, how high it rises, and how far it travels horizontally — starting from nothing but the two independent component equations of motion, for the ordinary case where the projectile lands back at the same height from which it was launched (so the net vertical displacement over the flight is zero).

Step 1. Write the vertical displacement as a function of time, taking upward positive: $y(t) = u\sin\theta\, t - \tfrac12 g t^2$ . This is just $s = ut + \tfrac12 at^2$ applied to the vertical direction alone, with $u_y = u\sin\theta$ and $a_y = -g$ .

Step 2. Find the time of flight $T$ by setting $y = 0$ , since the projectile returns to launch height at that instant. This gives $u\sin\theta\, T - \tfrac12 g T^2 = 0$ , or $T\left(u\sin\theta - \tfrac12 gT\right) = 0$ . The root $T=0$ is the launch instant itself; the physically meaningful root is $T = \dfrac{2u\sin\theta}{g}$ .

Step 3. Find the maximum height $H$ using the vertical velocity, $v_y = u\sin\theta - gt$ , which is zero exactly at the apex (the instant the projectile is momentarily neither rising nor falling). Solving $0 = u\sin\theta - gt_H$ gives $t_H = \dfrac{u\sin\theta}{g} = T/2$ — the rise time is exactly half the total flight time, because the fall retraces the rise for a level launch. Substituting $t_H$ into $y(t)$ from Step 1: $H = u\sin\theta\cdot\dfrac{u\sin\theta}{g} - \tfrac12 g\left(\dfrac{u\sin\theta}{g}\right)^2 = \dfrac{u^2\sin^2\theta}{2g}$ .

Step 4. Find the horizontal range $R$ using the horizontal motion, which never changes: $x(t) = u\cos\theta\, t$ . The range is simply this evaluated at $t = T$ : $R = u\cos\theta \cdot \dfrac{2u\sin\theta}{g} = \dfrac{2u^2\sin\theta\cos\theta}{g}$ .

Step 5. Simplify using the double-angle identity $2\sin\theta\cos\theta = \sin2\theta$ to reach the standard compact form $R = \dfrac{u^2\sin2\theta}{g}$ . Because $\sin2\theta$ is largest (equal to $1$ ) when $2\theta = 90^\circ$ , the range is maximum at $\theta = 45^\circ$ , giving $R_{max} = u^2/g$ ; and because $\sin2\theta = \sin\big(2(90^\circ-\theta)\big)$ , launch angles $\theta$ and $90^\circ - \theta$ always give the same range, arriving there by a low flat path and a high looping path respectively.

Worked examples

Board level - angled launch on level ground

A ball is thrown with a speed of $20\ \text{m/s}$ at an angle of $30^\circ$ above the horizontal. Take $g = 9.8\ \text{m/s}^2$ . Find (a) the time of flight, (b) the maximum height, and (c) the horizontal range.

Step 1. List the knowns and resolve the launch velocity: $u = 20\ \text{m/s}$ , $\theta = 30^\circ$ , so $\sin30^\circ = 0.5$ , $\cos30^\circ \approx 0.866$ , and (for the range formula) $\sin60^\circ \approx 0.866$ .

Step 2. Time of flight, from $T = \dfrac{2u\sin\theta}{g}$ : $T = \dfrac{2(20)(0.5)}{9.8} = \dfrac{20}{9.8} \approx \mathbf{2.04\ s}$ .

Step 3. Maximum height, from $H = \dfrac{u^2\sin^2\theta}{2g}$ : $H = \dfrac{(20)^2(0.5)^2}{2(9.8)} = \dfrac{400\times0.25}{19.6} = \dfrac{100}{19.6} \approx \mathbf{5.10\ m}$ .

Step 4. Horizontal range, from $R = \dfrac{u^2\sin2\theta}{g}$ : $R = \dfrac{(20)^2\sin60^\circ}{9.8} = \dfrac{400\times0.866}{9.8} \approx \dfrac{346.4}{9.8} \approx \mathbf{35.35\ m}$ .

Competitive level - horizontal launch from a tower, with a speed-at-landing extension

A stone is projected horizontally with speed $15\ \text{m/s}$ from the top of a tower $20\ \text{m}$ high. Take $g = 9.8\ \text{m/s}^2$ . Find (a) the time to hit the ground, (b) the horizontal distance from the base of the tower where it lands, and (c) the speed with which it strikes the ground.

Step 1. Recognise the special case: $\theta = 0$ , so $u_x = 15\ \text{m/s}$ and $u_y = 0$ ; the standard $T = 2u\sin\theta/g$ formula cannot be used because launch and landing heights differ by $20\ \text{m}$ , so the vertical equation must be solved directly instead.

Step 2. Vertical motion (taking downward positive for convenience here, starting from rest vertically): the fall height is $20\ \text{m} = \tfrac12 g t^2$ , so $t^2 = \dfrac{2(20)}{9.8} = \dfrac{40}{9.8} \approx 4.08$ , giving $t \approx \sqrt{4.08} \approx \mathbf{2.02\ s}$ .

Step 3. Horizontal distance, using the fact that $a_x = 0$ so the horizontal velocity never changes: $x = u_x t = 15 \times 2.02 \approx \mathbf{30.3\ m}$ .

Step 4. Landing speed, combining both components at impact: the vertical speed at impact is $v_y = gt = 9.8 \times 2.02 \approx 19.8\ \text{m/s}$ , while $v_x = 15\ \text{m/s}$ unchanged. The resultant speed is $v = \sqrt{v_x^2+v_y^2} = \sqrt{15^2 + 19.8^2} = \sqrt{225+392} = \sqrt{617} \approx \mathbf{24.8\ m/s}$ . Timing note: in an exam, spot the "projected horizontally from a height" phrasing immediately — it is the signal to set $\theta = 0$ in your head and switch straight to $h = \tfrac12 g t^2$ for time, rather than wasting time hunting for a way to force the level-ground $T$ , $H$ , $R$ formulas onto an unequal-height problem; the resultant-speed step in part (c) is a frequent extra mark that many students skip simply because they stop at horizontal distance.

Common traps

Using the standard $T$ , $H$ , $R$ formulas when launch and landing heights are different (a cliff, a tower, a raised platform) — those three formulas are derived assuming $y=0$ at both ends of the flight; unequal heights require solving $y = u_y t - \tfrac12 gt^2$ directly, as in the second worked example.
Forgetting that the horizontal velocity $u_x = u\cos\theta$ stays exactly constant throughout the flight, because $a_x = 0$ — a very common error is to apply deceleration or "slowing down" reasoning to the horizontal component the way you would to the vertical one.
Mixing up $\sin\theta$ and $\cos\theta$ when resolving the initial velocity — $u_x$ (horizontal, adjacent to $\theta$ ) uses cosine, and $u_y$ (vertical, opposite to $\theta$ ) uses sine; drawing the small right-angled velocity triangle before writing anything down avoids this.
Assuming maximum range and maximum height happen at the same angle — range is maximum at $\theta = 45^\circ$ , while height keeps increasing with $\theta$ all the way up to $90^\circ$ (a straight vertical throw, where the range is actually zero).
Dropping the factor of $2$ in $2\sin\theta\cos\theta = \sin2\theta$ when deriving $R$ from first principles, or forgetting to double the rise time $t_H$ to get $T$ for a level launch.

Common trapProjectile motion is not free fall with sideways drift — the two directions are genuinely independent, computed separately, and only combined again at the very last step

It is tempting to think of a projectile as "falling while also moving sideways," as if the sideways motion somehow slows the fall or the fall somehow curves the sideways path. Neither is true. Gravity has zero component along the horizontal, so the horizontal motion is completely unaffected by the fact that the object is also falling — a ball thrown horizontally and a ball simply dropped from the same height, at the same instant, hit the ground at exactly the same time, regardless of how fast the first one was thrown sideways. Solve the horizontal problem and the vertical problem as two separate, unrelated one-dimensional problems that merely share the same value of $t$ , and only combine $x(t)$ and $y(t)$ (or $v_x$ and $v_y$ ) together at the final step, such as when finding the resultant landing speed.

You should be able to…

Resolve a launch velocity $u$ at angle $\theta$ into horizontal and vertical components, $u\cos\theta$ and $u\sin\theta$ , without confusing sine and cosine.
Explain, in words, why the horizontal and vertical motions of a projectile are independent, and state the acceleration components $a_x = 0$ , $a_y = -g$ .
Derive the trajectory equation $y = x\tan\theta - \dfrac{gx^2}{2u^2\cos^2\theta}$ by eliminating time between the two component displacement equations.
Derive and apply $T = \dfrac{2u\sin\theta}{g}$ , $H = \dfrac{u^2\sin^2\theta}{2g}$ , and $R = \dfrac{u^2\sin2\theta}{g}$ for a level-ground launch, and explain why $R$ is maximum at $45^\circ$ .
Solve projectile problems where launch and landing heights differ — a horizontal launch from a tower or cliff — by solving the vertical equation of motion directly rather than forcing the level-ground formulas.
Combine the horizontal and vertical velocity components at any instant, using $v = \sqrt{v_x^2+v_y^2}$ , to find the resultant speed and direction of motion at that point.

Ask the tutor: Working through a projectile problem with an unusual launch condition — a cliff, an incline, or a landing speed and direction — ask the AI tutor to help you set up which equation to solve first.

Why does horizontal velocity stay constant when the ball is clearly slowing down as it rises?

It is the vertical velocity that slows as the ball rises, not the horizontal one — and it is easy to blur the two together because you are watching one moving object. Gravity is a purely vertical force here (air resistance is neglected), so by Newton's laws it can only change the vertical component of velocity; it has literally nothing to push against horizontally. If you watched only the ball's shadow on the ground moving with the sun directly overhead, that shadow would move at a perfectly steady speed, $u\cos\theta$ , the entire time the ball is in the air, even while the ball itself visibly rises and falls above it.

Why is the time to reach maximum height exactly half the total time of flight?

This is true only because the launch and landing are at the same height — the flight is symmetric in time about the apex. The vertical motion going up from $0$ to $u\sin\theta$ at $v_y=0$ under a constant deceleration $g$ is the mirror image, run backwards, of the vertical motion falling from $v_y=0$ back down to $-u\sin\theta$ under the same constant acceleration $g$ . Since gravity does not change during the flight, the time to slow from $u\sin\theta$ to $0$ equals the time to speed back up from $0$ to $u\sin\theta$ in the opposite direction, so each half takes $T/2$ . This symmetry breaks the moment landing height differs from launch height, which is exactly why the second worked example needed a direct calculation instead.

Does the trajectory equation assume the projectile lands at the same height it was launched from?

No — the trajectory equation $y = x\tan\theta - \dfrac{gx^2}{2u^2\cos^2\theta}$ is the full shape of the path itself, valid at every point along the flight regardless of where it eventually lands; it makes no assumption about landing height at all. It is only the three derived results, $T$ , $H$ , and $R$ , that assume level ground, because they were obtained by specifically setting $y=0$ to find where the path crosses back to launch height. If a tower or cliff is involved, you can still use the trajectory equation to find $y$ at any given $x$ , or work with $x(t)$ and $y(t)$ separately as in the second worked example.

Is projectile motion with air resistance ever going to come up on the CBSE exam?

Not at Class 11 — air resistance is explicitly neglected throughout this chapter, and every formula here, the trajectory, $T$ , $H$ , and $R$ , assumes a vacuum-like flight where gravity is the only force. Real thrown balls and shells do experience air drag, which shortens the range, lowers the maximum height, and skews the path away from a perfect symmetric parabola (the descending half becomes steeper than the ascending half), but the mathematics for that — a differential equation with a velocity-dependent drag force — is well beyond school algebra and is a university-level topic. For every board and competitive-exam problem at this level, treat the projectile as moving through a perfect vacuum near the Earth's surface.