JEE Mains Physics · Motion in a Plane

Projectile: Velocity, Trajectory and Projection from a Height

During the flight the horizontal velocity u cos θ never changes while the vertical velocity falls by g every second; the path is the parabola y = x tan θ − gx²/(2u² cos²θ), and a body thrown horizontally from a height h falls for √(2h/g) whatever its speed.

All Motion in a Plane notes JEE Mains Physics strategy

Why this matters

Twenty-four PYQs, seventeen of them multiple choice, and three from 2026. Thirteen ask about the state of the projectile at one instant: its speed, its direction, its height at a given time, or its kinetic energy at the top. Four give the equation of the path and seven throw the body from a height, horizontally or down a stairway. Each of them starts from the same fact: the horizontal velocity is the same at every point of the flight.

Concept 1 of 3: Velocity and kinetic energy during the flight

The horizontal velocity u cos θ is the same at launch, at the top and at landing. Only the vertical velocity changes: it falls by g every second, is zero at the top, and comes back to −u sin θ on landing. So at the top the speed is u cos θ, not zero, and the kinetic energy there is the launch value times cos²θ. A projectile passes each height twice, at two times placed symmetrically about the top.

Definition

$v_x = u\cos\theta$ throughout; $v_y = u\sin\theta - gt$ .
Speed $v = \sqrt{v_x^{2} + v_y^{2}}$ ; direction $\tan\beta = \dfrac{u\sin\theta - gt}{u\cos\theta}$ above the horizontal.
At the top: speed $u\cos\theta$ , kinetic energy $K\cos^{2}\theta$ where K is the launch value; potential energy is greatest there.
Height at time t: $y = u\sin\theta\,t - \tfrac{1}{2}gt^{2}$ .
Same height at $t_1$ and $t_2$ : $T = t_1 + t_2$ and $h = \tfrac{1}{2}gt_1t_2$ .
Change in momentum from launch to landing: $2mu\sin\theta$ , straight down.

Velocity during the flight

v_x = u\cos\theta,\ v_y = u\sin\theta - gt \qquad K_{top} = K\cos^{2}\theta \qquad h = \tfrac{1}{2}gt_1t_2

Worked example

A ball is thrown at 50 m/s at

53^{\circ}

to the horizontal

(\sin 53^{\circ} = 0.8,\ g = 10\ \text{m/s}^{2})

. Find its velocity 2 s after launch, and the fraction of its launch kinetic energy it has at the top.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 1 · Q7Moderate

Example 1 · Motion in a Plane · Projectile: Velocity, Trajectory and Projection from a Height

A projectile is thrown upward at an angle

60^{\circ}

with the horizontal. The speed of the projectile is

20\text{ }m/s

when its direction of motion is

45^{\circ}

with the horizontal. The initial speed of the projectile is

\_\_\_\_

m/s

Zero speed at the top

Only the vertical velocity is zero at the top. The horizontal velocity u cos θ is still there, so the speed and the kinetic energy are not zero.

cos θ instead of cos²θ

Kinetic energy goes as speed squared. The speed at the top is u cos θ, so the kinetic energy there is K cos²θ: at 37° (cos 37° = 0.8) that is 0.64K, not 0.8K.

Drill 12 more on velocity and kinetic energy during the flight

Concept 2 of 3: Equation of the trajectory

Eliminate t between x = u cos θ t and y = u sin θ t − ½gt², and the path comes out as y = x tan θ − (a constant) x²: a parabola opening downwards. Read it backwards when the question gives the path: the coefficient of x is the slope at launch, tan θ, and the coefficient of x² fixes the horizontal speed. The range and the height follow from the same two numbers.

Definition

$y = x\tan\theta - \dfrac{gx^{2}}{2u^{2}\cos^{2}\theta} = x\tan\theta\left(1 - \dfrac{x}{R}\right)$ .
Given $y = \alpha x - \beta x^{2}$ : $\tan\theta = \alpha$ , $R = \dfrac{\alpha}{\beta}$ , $H = \dfrac{\alpha^{2}}{4\beta}$ (at $x = R/2$ ), and $\beta = \dfrac{g}{2u_x^{2}}$ .
The vertical launch speed is $u_y = \alpha u_x$ .
A point (x, y) on the path: substitute it into the equation to find u or θ.

Path of a projectile

y = x\tan\theta - \frac{gx^{2}}{2u^{2}\cos^{2}\theta} \qquad y = \alpha x - \beta x^{2}:\ R = \frac{\alpha}{\beta},\ H = \frac{\alpha^{2}}{4\beta}

Worked example

A projectile follows

y = 2x - \dfrac{x^{2}}{20}

(x, y in metres,

g = 10\ \text{m/s}^{2}

). Find the angle of projection, the range, the greatest height and the launch speed.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 27 July 2022 · Q92Moderate

Example 2 · Motion in a Plane · Projectile: Velocity, Trajectory and Projection from a Height

A body of mass

10\text{ }kg

is projected at an angle of

45^{\circ}

with the horizontal. The trajectory of the body is observed to pass through a point

(20,10)

. If

T

is the time of flight, then its momentum vector, at time

t =\frac{T}{\sqrt{2}}

, is [Take

g = 10\text{ }m/s^{2}

]

Dropping the cos²θ

The x² coefficient is g/(2u² cos²θ) = g/(2uₓ²): it uses the horizontal speed, not the launch speed. Using g/(2u²) gives the wrong speed.

The top is at x = R/2

The greatest height comes halfway across. Setting dy/dx = 0 gives x = α/2β, which is R/2; substituting gives H = α²/4β.

Drill 3 more on equation of the trajectory

Concept 3 of 3: Thrown horizontally from a height

A body thrown horizontally starts with no vertical velocity, so it falls exactly as a dropped body does: it takes √(2h/g) to reach the ground, however fast it is thrown. The horizontal speed only decides how far out it lands. A packet dropped from a plane is the same problem, with the plane's speed as the horizontal speed.

Definition

Time of fall $t = \sqrt{2h/g}$ ; horizontal distance $x = u\sqrt{2h/g}$ .
Landing velocity: $v_x = u$ , $v_y = \sqrt{2gh}$ , $v = \sqrt{u^{2} + 2gh}$ , at $\tan^{-1}\dfrac{\sqrt{2gh}}{u}$ below the horizontal.
Distance from the release point to the landing point: $\sqrt{x^{2} + h^{2}}$ .
Stairway, steps of height and width d: the ball reaches the level of step n at $x = u\sqrt{2nd/g}$ and lands on step n if $x \le nd$ . The least speed to reach step n just clears the corner of step n − 1: $u^{2} = \dfrac{g(n - 1)d}{2}$ .
A body that falls from H and is turned horizontal at height h (speed kept) spends $\sqrt{2(H - h)/g} + \sqrt{2h/g}$ in the air, greatest at $h = H/2$ .
'Thrown at an angle from a tower' may be above or below the horizontal: solve $-h = u_yt - \tfrac{1}{2}gt^{2}$ with the sign of $u_y$ set by the stem.

Horizontal projection from height h

t = \sqrt{\frac{2h}{g}} \qquad x = u\sqrt{\frac{2h}{g}} \qquad v = \sqrt{u^{2} + 2gh}

Worked example

A stone is thrown horizontally at 15 m/s from the top of a 45 m tower

(g = 10\ \text{m/s}^{2})

. When and where does it land, and with what velocity?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 7 Apr 2025 · Q93Moderate

Example 3 · Motion in a Plane · Projectile: Velocity, Trajectory and Projection from a Height

A helicopter flying horizontally with a speed of

360\text{ }km/h

at an altitude of 2 km , drops an object at an instant. The object hits the ground at a point O , 20 s after it is dropped. Displacement of 'O' from the position of helicopter where the object was released is :(use acceleration due to gravity

g= 10\text{ }m/s^{2}

and neglect air resistance)

Giving the horizontally thrown body a vertical speed

Thrown horizontally means v_y = 0 at release. The time of fall is √(2h/g), the same as dropping it; a faster throw lands farther out but not sooner.

Horizontal distance is not displacement

The landing point is x out and h down from the release point. The displacement is √(x² + h²); the horizontal distance x alone is a common wrong option.

Drill 6 more on thrown horizontally from a height

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Velocity and kinetic energy during the flight
Velocity during the flight
$v_x = u\cos\theta,\ v_y = u\sin\theta - gt \qquad K_{top} = K\cos^{2}\theta \qquad h = \tfrac{1}{2}gt_1t_2$
Equation of the trajectory
Path of a projectile
$y = x\tan\theta - \frac{gx^{2}}{2u^{2}\cos^{2}\theta} \qquad y = \alpha x - \beta x^{2}:\ R = \frac{\alpha}{\beta},\ H = \frac{\alpha^{2}}{4\beta}$
Thrown horizontally from a height
Horizontal projection from height h
$t = \sqrt{\frac{2h}{g}} \qquad x = u\sqrt{\frac{2h}{g}} \qquad v = \sqrt{u^{2} + 2gh}$

Watch out for (6)

Zero speed at the top
→ Velocity and kinetic energy during the flight
cos θ instead of cos²θ
→ Velocity and kinetic energy during the flight
Dropping the cos²θ
→ Equation of the trajectory
The top is at x = R/2
→ Equation of the trajectory
Giving the horizontally thrown body a vertical speed
→ Thrown horizontally from a height
Horizontal distance is not displacement
→ Thrown horizontally from a height

Test yourself on Motion in a Plane

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