Q.An object is thrown at angle θ with horizontal. Show that the path followed by the object (Trajectory) is parabolic.
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Projectile Motion
A projectile moves under gravity alone, and the one idea that solves every problem is that its motion splits into two independent one-dimensional motions: horizontal at constant velocity (no force), and vertical at constant acceleration −g. Handle each axis separately with the equations of motion; the only thing they share is the time t. (Take g = 10 m/s² unless a problem says otherwise.)
1 — Ground-to-ground launch at angle θ, speed u. With uₓ = u cosθ, u_y = u sinθ:
- Time of flight:
T = 2u sinθ / g - Maximum height:
H = u²sin²θ / 2g - Range:
R = u²sin2θ / gThe range is maximum at θ = 45° (R_max = u²/g, whereH = R_max/4); two complementary anglesθand90°−θgive the same range. Useful relations:R = 4H/tanθ(sotanθ = 4H/R) andH = gT²/8.
2 — Velocity anywhere. The horizontal velocity uₓ = u cosθ never changes; the vertical velocity is v_y = u sinθ − gt. So the speed is √(uₓ² + v_y²) and the direction is tan φ = v_y/uₓ. At the highest point v_y = 0, so the velocity is purely horizontal = u cosθ (not zero — a classic trap), while the acceleration is still g downward. At a height h, v_y² = (u sinθ)² − 2gh; on returning to the launch level the speed equals the launch speed u.
3 — The trajectory is a parabola: y = x tanθ − gx²/(2u²cos²θ) = x tanθ (1 − x/R). Reading y = ax − bx² gives tanθ = a and R = a/b. This is also the tool for clearing a wall of height h at distance d, or hitting a target (x, y) — substitute and solve.
4 — Horizontal projectile from a height h (launched horizontally at speed u): the fall time t = √(2h/g) is independent of u; the horizontal range is x = u·t; the vertical speed on landing is v_y = √(2gh), the landing speed √(u² + 2gh), at angle tan φ = v_y/u below the horizontal. Launched from a height at an angle, set the vertical displacement to −h and solve the quadratic for the flight time.
5 — Projectile on an incline (angle α). Take axes along and perpendicular to the incline; gravity splits into g sinα (along, down-slope) and g cosα (perpendicular). Time of flight up the incline = 2u sin(θ−α)/(g cosα), and the range up the incline is maximum at θ = 45° + α/2. …
Resolving the initial velocity into horizontal and vertical components and eliminating time t between the two equations of motion gives an equation of the form y = x tanθ − (g x²)/(2u²cos²θ), which is the equation of a parabola. …
Eliminating time between the horizontal and vertical equations of motion for a projectile gives y = Ax − Bx², the equation of a parabola.
Consider an object projected with initial speed u at angle θ above the horizontal, from the origin, under gravity g (taking x horizontal, y vertical, and air resistance neglected).
Initial velocity components:
ux=ucosθ (constant, no horizontal acceleration)
uy=usinθ
Horizontal motion (uniform velocity):
x=ucosθ⋅t⇒t=ucosθx
Vertical motion (uniform acceleration −g):
y=usinθ⋅t−21gt2
Substitute the expression for t from the horizontal equation into the vertical equation:
y=usinθ⋅ucosθx−21g(ucosθx)2
y=xtanθ−2u2cos2θgx2
…
Showing the 12 most recent of 13 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.The shape of trajectory of the motion of an object is determined by(a) acceleration(b) initial position(c) initial velocity(d) all of these
›Reveal solutionSolution
To fully determine a particle's trajectory (path in space), you need all three pieces of initial-condition information together: where it starts, how fast/which way it starts moving, and what acceleration acts on it throughout.
Given a constant acceleration a, the position at time t is found from:
r(t) = r0 + v0 t + (1/2) a t^2
- Initial position r0 fixes WHERE the path is located in space (translates the whole curve).
- Initial velocity v0 fixes the direction the object starts moving in, which combines with the acceleration direction to decide whether the path is a straight line (v0 parallel/antiparallel to a) or curved, e.g. a parabola (v0 not parallel to a). …
- CBSE 2026Set ANNUAL1 markMCQQ.A projectile is launched horizontally from a certain height. What is the shape of its trajectory?(a) Straight line(b) Parabola(c) Circle(d) Ellipse
›Reveal solutionSolution
Horizontal launch = horizontal velocity component constant, vertical motion accelerates under gravity -- combining a linear (x) relation with a quadratic (y) relation in time always produces a parabola.
Let the object be launched horizontally with speed u from height h.
- Horizontal: x = u t (no horizontal acceleration)
- Vertical: y = (1/2) g t^2 (falling under gravity, starting with zero vertical velocity) …
- CBSE 2025Set ANNUAL1 markMCQQ.The directions of velocity and acceleration at the top of the trajectory of a projectile are(a) parallel to each other(b) opposite to each other(c) at an angle of 45 degrees to each other(d) perpendicular to each other
›Reveal solutionSolution
At the highest point of a projectile's path, the vertical velocity component is momentarily zero, leaving only the horizontal component -- while gravity's acceleration always points straight down, so the two vectors are at 90 degrees.
For a projectile launched at an angle, the vertical component of velocity (v_y) decreases due to gravity, becomes zero exactly at the top of the trajectory, and then reverses sign on the way down.
At the topmost point: v_y = 0, so the velocity vector is entirely horizontal (v = v_x, in the horizontal direction).
…
- CBSE 2025Set ANNUAL1 markMCQQ.If a projectile is thrown with velocity u at an angle θ with the horizontal, then the velocity at maximum height during the projectile motion will be:(a) u sin θ(b) u cos θ(c) 2u sin θ(d) 2u cos θ
›Reveal solutionSolution
At the top of a projectile's trajectory, the vertical velocity is momentarily zero (that is why it's the highest point) — only the constant horizontal component ucosθ remains.
A projectile launched with speed u at angle θ to the horizontal has two independent components of velocity:
- Horizontal: ux=ucosθ (constant throughout the flight, since there is no horizontal force under gravity alone)
- Vertical: uy=usinθ, which decreases under gravity as vy=usinθ−gt …
- CBSE 2025Set ANNUAL1 markMCQQ.The horizontal range of projectile is maximum when the angle of projectile is-(a) 45°(b) 60°(c) 30°(d) 0°
›Reveal solutionSolution
The range of a projectile is maximum at a projection angle of 45°.
The horizontal range of a projectile launched with speed u at angle θ to the horizontal is:
R=gu2sin(2θ)
…
- CBSE 2025Set hz1 markMCQQ.At what point in its trajectory does a projectile have minimum upward speed.(a) At the start of the Trajectory(b) At the end of the Trajectory(c) At the midpoint of the Trajectory(d) At the highest point in its Trajectory
›Reveal solutionSolution
The upward (vertical) speed of a projectile keeps decreasing due to gravity and becomes zero exactly at the highest point of the trajectory.
For a projectile launched with initial speed u at angle theta to the horizontal, the vertical component of velocity at time t is:
v_y = u sin(theta) - g t
At the start (t = 0), v_y = u sin(theta), which is the maximum upward speed.
…
- CBSE 2025Set sz1 markMCQQ.If a projectile is thrown with velocity u at an angle theta with the horizontal, then velocity at maximum height during projectile motion will be: (A) u sin theta (B) 2u sin theta (C) 2u cos theta (D) u cos theta
›Reveal solutionSolution
At maximum height the vertical velocity is zero and only the horizontal component u cos(theta) remains.
The initial velocity u makes angle theta with the horizontal, so its components are ucosθ (horizontal) and usinθ (vertical). …
- CBSE 2024Set ANNUAL1 markMCQQ.In projectile motion, the horizontal component of velocity remains(a) constant(b) increasing(c) decreasing(d) zero
›Reveal solutionSolution
Gravity only acts vertically (downward), so it never changes the horizontal velocity component — it stays constant throughout the projectile's flight.
In projectile motion (ignoring air resistance), the only force acting is gravity, which is purely vertical (g, downward). Newton's second law tells us acceleration only exists along the direction of net force, so: …
- CBSE 2024Set ANNUAL1 markMCQQ.In a projectile motion, the horizontal range is maximum for angle of projection(a) a) 0°(b) b) 45°(c) c) 60°(d) d) 90°
›Reveal solutionSolution
[!TLDR]
b) 45°
Why
Range R = u2 sin(2θ)/g is maximum when sin(2θ)=1, i.e. …
- CBSE 2022Set TERM11 markMCQQ.Which of the following remains constant for a projectile motion?(1) Kinetic energy(2) Momentum(3) Horizontal component of velocity(4) Vertical component of velocity
›Reveal solutionSolution
The only force on a projectile is gravity, acting vertically downward. With zero horizontal force, the horizontal velocity component never changes -- everything else (KE, momentum, vertical velocity) changes as the projectile rises and falls.
In projectile motion, the acceleration due to gravity, g, acts purely in the vertical direction. There is no horizontal acceleration.
- Horizontal component of velocity, v_x = u cos(theta): remains CONSTANT throughout the flight, since a_x = 0. …
- CBSE 2022Set TERM11 markMCQQ.In case of a projectile motion, what is the angle between the velocity and acceleration at the highest point?(1) 0°(2) 45°(3) 90°(4) 180°
›Reveal solutionSolution
At the top of the trajectory the vertical velocity component is momentarily zero, leaving only the horizontal velocity component -- which is perpendicular to the ever-present, purely vertical, gravitational acceleration.
At the highest point of a projectile's path, the vertical component of velocity, v_y = u sin(theta) - gt, has dropped to zero (the projectile is momentarily neither rising nor falling). What remains is purely the horizontal component, v_x = u cos(theta), so the velocity vector at that instant points horizontally.
…
- CBSE 2022Set sz1 markQ.At which angle, the height attained by a projectile is maximum?
›Reveal solutionSolution
Maximum height H = u^2 sin^2(theta) / (2g) is largest when sin(theta) = 1, i.e. theta = 90 degrees.
For a projectile launched with speed u at angle theta to the horizontal, the maximum height reached is:
H = u^2 sin^2(theta) / (2g)
For a fixed launch speed u, H depends only on sin^2(theta). This is maximum when sin(theta) = 1, which occurs at theta = 90 degrees. At this angle the projectile has no horizontal component of velocity at all -- it is thrown straight up, so its entire initial speed contributes to gaining height.
…
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.