Projectile Motion

The set-up and the master idea

NCERT §3.9 defines a projectile as any object that is in flight after being projected and is acted upon only by gravity. A football, a bullet, a relief packet dropped from a plane — all are projectiles. Air resistance is neglected and $g$ is taken constant.

Place the origin at the launch point. Let the projectile be launched with speed $v_0$ at angle $\theta_0$ above the horizontal. Resolve the initial velocity:

The acceleration throughout the flight is purely vertical and downward:

Because gravity has no horizontal component, the horizontal motion proceeds at constant velocity. The vertical motion is identical to the free fall of a body launched straight up with speed $v_{0y}$. Galileo's principle of independence (Dialogue on the Great World Systems, 1632) says these two one-dimensional motions can be solved separately and recombined at the end.

NIOS §4.1 makes this visible with a balcony experiment: drop one ball and, at the same instant, throw a second ball horizontally off the edge. Both hit the ground at exactly the same moment. The thrown ball's vertical motion is unaffected by its horizontal speed.

Position and velocity equations

Setting $x_0 = y_0 = 0$ at the launch point and using $a_x = 0$, $a_y = -g$, NCERT Eq. (3.37) and (3.38) give the four equations that govern the projectile:

Three things to internalise. $v_x$ has no $t$ — the horizontal component is locked in at launch. $y(t)$ is the same as a body launched straight up with speed $v_0\sin\theta_0$. At $t = 0$ both $x$ and $y$ are zero. The rest of the subtopic extracts consequences from these four equations.

The parabolic trajectory

What shape does the path trace? Eliminate $t$. From the $x$ equation, $t = x / (v_0\cos\theta_0)$. Substitute into the $y$ equation:

This is a parabola of the form $y = ax - bx^{2}$, with $a = \tan\theta_0$ positive (the path rises) and $b = g/(2 v_0^{2}\cos^{2}\theta_0)$ positive (the path bends back down). The path of any projectile under uniform gravity and zero air drag is a parabola — NCERT Eq. (3.39), NIOS Eq. (4.8b).

Time of flight

The time of flight $T$ is the total time the projectile spends in the air, measured from launch to the instant it returns to the launch height. Setting $y = 0$ in $y(t) = (v_0\sin\theta_0)t - \tfrac{1}{2}g t^{2}$ gives two roots: $t = 0$ (launch) and the one we want,

The time to reach maximum height — at which $v_y = 0$ — is half of this:

The symmetry $T = 2 t_m$ is a consequence of the parabola being symmetric about its peak. Going up and coming down to the same level take equal times. This is NCERT Eq. (3.40a, b).

Maximum height

At the peak, the vertical velocity is zero. Using $v_y^{2} = v_{0y}^{2} - 2 g\,y$ with $v_y = 0$ and $v_{0y} = v_0\sin\theta_0$,

This is NCERT Eq. (3.41). Two NEET-relevant features. First, $H$ depends only on the vertical component $v_0\sin\theta_0$ of the launch velocity — the horizontal speed never enters. Two projectiles launched with different angles but the same vertical component reach the same peak height. Second, $H$ scales as $v_0^{2}$, so doubling the launch speed quadruples the height (for fixed angle).

Horizontal range and the 45° optimum

The horizontal range $R$ is the horizontal distance covered between launch and the return to launch height. It equals the constant horizontal velocity multiplied by the time of flight:

Using the identity $2\sin\theta\cos\theta = \sin 2\theta$,

For a fixed launch speed $v_0$, the range is largest when $\sin 2\theta_0$ is largest. The maximum value of the sine function is $1$, achieved at $2\theta_0 = 90°$, i.e. $\theta_0 = 45°$. At this angle,

So a projectile achieves its longest horizontal flight when launched at 45° above the horizontal. NCERT Eq. (3.42a, b). NIOS §4.1.1 records the same result and notes that long-jumpers, hammer-throwers and javelin-throwers all approximate this angle in practice — though real-world athletes use slightly less because they are not launching from ground level.

The complementary-angle trap (NEET favourite)

This is the single most-tested subtlety in projectile motion. NCERT Example 3.6 records it; Galileo proved it in 1632. Take two launch angles that are complementary, that is, they add up to $90°$:

For example $30°$ and $60°$, or $40°$ and $50°$, or $15°$ and $75°$. The ranges are identical, because

So launching at $30°$ and $60°$ with the same $v_0$ deposits the projectile at exactly the same spot on level ground. The trap is that students conclude the two motions are otherwise identical. They are not. The steeper launch climbs higher and stays in the air longer.

Velocity at any instant

At time $t$ the velocity components are $v_x = v_0\cos\theta_0$ and $v_y = v_0\sin\theta_0 - g t$. The instantaneous speed is

and the direction with the horizontal is given by $\tan\alpha = v_y / v_x$. Two NEET-favourite checkpoints follow directly.

At the peak. At maximum height $v_y = 0$, so the speed is

The projectile is moving purely horizontally and the speed is not zero — it is the horizontal component of the launch velocity. This is the single most common distractor in NEET. The 2022 paper made the question even more pointed by stating the angle "with the vertical direction" so the candidate has to translate it to $60°$ with the horizontal before applying $v_0\cos 60° = 10 \times 0.5 = 5$ m s$^{-1}$. (The official answer is $5\sqrt{3}$ because the stem gave $60°$ from the vertical, leaving $30°$ with the horizontal and $v_0\cos 30° = 5\sqrt{3}$.)

At impact on level ground. At the landing instant $t = T$, we have $v_y = v_0\sin\theta_0 - g T = -v_0\sin\theta_0$. The vertical component is equal in magnitude and opposite in sign to its launch value. The horizontal component is unchanged. Therefore the landing speed equals the launch speed and the angle below the horizontal equals the launch angle above. This is the projectile's energy-conservation signature.

R, H, T shortcut relations NEET rewards

Several derived identities trim the algebra in NEET problems. Memorise these — they are time-savers, not luxuries.

• R in terms of H and angle. Dividing $R$ by $H$ gives $R/H = 4\cot\theta_0$, i.e. $R = 4 H \cot\theta_0$. So if you know any two of $R$, $H$, $\theta_0$, you can find the third without touching $v_0$.
• Maximum range identity. $R\,g = v_0^{2}\sin 2\theta_0$. At $\theta_0 = 45°$ this gives $R_{\max}\,g = v_0^{2}$.
• Complementary heights. If $H_1$ and $H_2$ are the maximum heights for two complementary launches with the same $v_0$, then $H_1 + H_2 = v_0^{2}/(2g)$. Proof: $H_1 + H_2 = v_0^{2}(\sin^{2}\theta + \sin^{2}(90°-\theta))/(2g) = v_0^{2}(\sin^{2}\theta + \cos^{2}\theta)/(2g) = v_0^{2}/(2g)$.
• Complementary times. $T_1\cdot T_2 = (2 v_0\sin\theta/g)\cdot(2 v_0\cos\theta/g) = 4 v_0^{2}\sin\theta\cos\theta/g^{2} = 2 R/g$.

Horizontal launch from a height

NCERT Example 3.7 (hiker on a cliff) and NIOS §4.1 cover this case directly. Launch from height $h$ with horizontal speed $u$ and $v_{0y} = 0$. Place the origin at the launch point with $y$ measured downward as positive (or upward as positive with $y_{\text{ground}} = -h$). The ball hits the ground when $\tfrac{1}{2}g t^{2} = h$, giving

The time of fall is identical to that of a body dropped from rest at the same height — direct confirmation of the independence principle. The 2021 NEET PYQ Q.44 (car-and-ball) is exactly this setup: at the instant of release the ball acquires the car's speed as its horizontal launch component, and falls with $g$ vertically. The horizontal velocity of the ball is the car's velocity at the release instant, and stays unchanged.

Projectile on an inclined plane (brief)

NEET occasionally tests projectiles launched along or up an incline of angle $\alpha$. Rotate the axes so the new $x$-axis lies along the incline and the new $y$-axis is perpendicular to it. The gravitational acceleration then has components $-g\sin\alpha$ along the incline and $-g\cos\alpha$ perpendicular to it. With the projectile launched at angle $\theta$ to the incline and initial speed $u$, the time of flight (return to the incline) is

and the range up the incline is

The optimum launch angle for maximum range up the incline is $\theta = (90° - \alpha)/2$, not $45°$. Treat the inclined-plane case as bonus material — appearance frequency is roughly one paper in five.

Worked NEET-style examples