Projectile Motion
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Direct answer
Projectile motion is two-dimensional motion under gravity alone, analysed by resolving the initial velocity u at angle theta into independent components: the horizontal motion is uniform (no horizontal force) and the vertical motion is uniformly accelerated at g. The path is a parabola, with time of flight T = 2 u sin(theta)/g, maximum height H = u^2 sin^2(theta)/(2g) and horizontal range R = u^2 sin(2theta)/g for level-ground projection.
What you must remember
- Time of flight T = 2 u sin(theta)/g; time to the highest point is half of this.
- Maximum height H = u^2 sin^2(theta)/(2g); range R = u^2 sin(2theta)/g; range is maximum at 45 degrees, and complementary angles give equal ranges.
- Trajectory: y = x tan(theta) − g x^2/(2 u^2 cos^2(theta)), the equation of a parabola.
- At the highest point the velocity is horizontal, u cos(theta), and not zero; only the vertical component vanishes.
- Speed at any height follows from energy conservation: v^2 = u^2 − 2 g times the vertical displacement, so the projectile regains its launch speed on returning to the same level.
- Horizontal projection from height h: t = sqrt(2h/g), range = u sqrt(2h/g).
- For two projectiles released together, the relative acceleration is zero (both fall at g), so one appears to move in a straight line relative to the other.
Common confusion
The classic error is applying the level-ground range formula when landing and launch heights differ, as in a throw from a cliff or a ball struck at a window above the ground. The formula assumes equal heights; otherwise set up y = u sin(theta) t − (1/2) g t^2 with the actual vertical displacement and solve for time first. Students also wrongly take the top-of-flight velocity as zero, or forget that a body dropped from a moving vehicle keeps the vehicle's horizontal velocity.
Exam-focused takeaway
JEE Main tests T, H, R and the trajectory equation, plus drops from balloons or aeroplanes, as single-correct and numerical-value questions — quick marks if component bookkeeping is clean. JEE Advanced extends to inclined-plane projectiles (resolve g along and perpendicular to the incline), the envelope of safe trajectories, and pairs of particles projected toward each other where relative-velocity reasoning settles the separation or the time of closest approach. Treat every case as two independent one-dimensional motions, with relative motion as the Advanced-layer add-on.
Frequently asked questions
Why is the horizontal velocity constant?
Gravity acts vertically and no horizontal force acts (air resistance neglected), so the horizontal component u cos(theta) never changes.
What is the velocity at the highest point?
Horizontal and equal to u cos(theta); only the vertical component is zero there, so the speed is minimum but not zero.
At what angle is the range maximum?
45 degrees on level ground; complementary angles theta and (90 degrees − theta) give the same range.
Can the range formula be used when landing occurs at a different height?
No — it holds only for equal launch and landing levels; otherwise solve the vertical displacement equation for time and multiply by the horizontal velocity.
How is a projectile on an inclined plane handled?
Resolve g into g sin(theta) along and g cos(theta) perpendicular to the incline, then apply the usual kinematics in those two directions.
What happens to a ball dropped from a moving train?
It retains the train's horizontal velocity and falls in a parabola, appearing to fall straight down to a passenger on the train.