sci_phy

One Throw, Two Motions: Vectors and Projectiles

Chapter summary, hard words and model exam answers.

Free online summary and notes. Read it here, no PDF download needed.

About the author

Science · CBSE Class 11 · NCERT Physics Part I, Ch.3

Summary

Everything about motion so far, in Class 9 and in the previous chapter, lived on a single straight line, where 'direction' was never more complicated than a plus or minus sign. Real motion rarely cooperates with that restriction: a footballer's run curves, a thrown ball arcs, a satellite loops. Describing motion across a flat surface, in two directions at once, needs a more capable kind of quantity than a single signed number. A scalar, like distance, mass or temperature, is fully described by one number and a unit. A vector, like displacement, velocity, acceleration or force, needs a number and a direction together, and vectors combine using their own rules, not ordinary arithmetic: laying two vectors head to tail and drawing the line from the first tail to the last head gives their sum, a method that works regardless of how many directions are involved. A position in a plane can be described by a position vector, drawn from a chosen origin to the object's location, and if an object moves from one position to another, the straight-line vector connecting old position to new position, regardless of the actual path taken, is its displacement, the same displacement concept from Class 9, now genuinely two-dimensional.

Suppose rain is falling straight down at 8 m/s, and a gust of wind simultaneously blows sideways at 5 m/s. What is the actual velocity of the rain a person standing outside experiences? It is tempting to just add 8 and 5, but that only works if two things point the same way, and here they are at right angles. This is exactly the situation vectors are built for: rather than adding magnitudes directly, break each vector into components along two convenient perpendicular directions, usually horizontal (x) and vertical (y), using Aₓ = A cos θ and Aᵧ = A sin θ, where θ is the angle the vector makes with the x-axis. Components along the same axis add normally, since they are now genuinely parallel, so the resultant's components are just the sums of the individual x-components and y-components, and the resultant's own magnitude follows from these using the Pythagorean relation A = √(Aₓ² + Aᵧ²). For the falling rain, treating downward and sideways as the two perpendicular directions, the resultant speed is √(8² + 5²), which is √89, or about 9.4 m/s, at an angle of tan⁻¹(5/8), about 32°, from the vertical, which is exactly the angle a person should tilt an umbrella to stay properly dry.

The instantaneous velocity and acceleration built from limits in the previous chapter carry over directly to a plane, just resolved into two components instead of one: v = dr/dt, where r is the position vector, resolves into vₓ = dx/dt and vᵧ = dy/dt, and a = dv/dt resolves the same way into aₓ = dvₓ/dt and aᵧ = dvᵧ/dt. This has a genuinely powerful consequence: motion in a plane with constant acceleration can be treated as two completely separate one-dimensional motions happening at the same time, one along x, one along y, each obeying the familiar kinematic equations independently, with no interaction between them at all. Direction, in a plane, is always tangential: at any point along a curved path, the instantaneous velocity vector points exactly along the tangent to that path at that point, the direction the object is heading at that exact instant, whatever the shape of the path leading up to it.

A projectile, anything in flight after being launched, whether a football, a cricket ball, or a jet of water from a fountain, is the single richest application of everything so far, thanks to an insight Galileo first stated clearly: the horizontal and vertical parts of a projectile's motion are completely independent of each other. Launch an object at speed v₀ and angle θ₀ above the horizontal, and split that launch velocity into components immediately: vₓ = v₀ cos θ₀, vᵧ = v₀ sin θ₀. Once airborne, ignoring air resistance, the only force acting is gravity, straight down, so the horizontal component of velocity never changes at all throughout the flight, exactly like an object moving at constant velocity with nothing acting on it; meanwhile, the vertical component behaves exactly like a ball thrown straight up, decelerating under gravity, momentarily stopping at the peak, then accelerating back down. Neither motion knows or cares about the other; they simply happen simultaneously, and the object's actual position at any instant is wherever those two independent motions happen to place it. Eliminating time between the two component equations shows that the path traced out, y as a function of x, has the form y = (tan θ₀)x - [g/(2v₀²cos²θ₀)]x², an equation of a parabola, exactly the curved arc you actually see when you watch something get thrown.

Three practical questions about a projectile all follow directly from treating the vertical motion as free fall in disguise. How long is it airborne? The vertical velocity returns to zero at the peak, at time tₘ = v₀ sin θ₀ / g, and by the symmetry of the parabola, the total time of flight until landing back at launch height is exactly double that: T = 2v₀ sin θ₀ / g. How high does it rise? Substituting tₘ into the vertical position equation gives the maximum height, h = (v₀ sin θ₀ )² / 2g. How far does it travel horizontally before landing? Multiplying the constant horizontal velocity by the total time of flight gives the range, R = v₀² sin 2θ₀ / g, which is largest when sin 2θ₀ is largest, that is, when θ₀ is exactly 45°. Consider a volleyball served at 18 m/s at 40° above the horizontal, with g = 9.8 m/s²: its time of flight is 2(18)(sin 40°)/9.8, about 2.36 s; its maximum height is (18 sin 40°)² / (2 x 9.8), about 6.8 m; and its range is (18)² sin 80° / 9.8, about 32.6 m. Every one of these numbers comes from nothing more exotic than free fall and constant horizontal velocity, combined.

Class 9 established that an object moving in a circle at constant speed is still accelerating, because its direction keeps changing. Now that direction is a proper vector quantity, exactly where that acceleration points, and how large it is, can be pinned down. As an object moves a small angle around a circle of radius R, its velocity vector, always tangent to the circle, rotates by that same small angle. The change in velocity this produces, worked out geometrically, turns out to point directly inward, towards the centre of the circle, at every single point along the path, never outward, never sideways. This inward-pointing acceleration is called centripetal acceleration, from Latin for 'centre-seeking', and its magnitude works out to aᴄ = v²/R. A second way to describe the same motion uses angular speed ω, the rate at which the angle around the circle changes, related to ordinary speed by v = ωR, which lets centripetal acceleration be written equivalently as aᴄ = ω²R. If the object completes one full circle in time T (its time period), with frequency ν = 1/T revolutions per second, then ω = 2πν and v = 2πRν, connecting every version of circular motion, however it happens to be described, back to the same single inward-pointing acceleration.

Hard words & meanings

scalara physical quantity fully specified by a magnitude alone
vectora physical quantity specified by both a magnitude and a direction
resolving a vectorsplitting a vector into components along two (or three) chosen perpendicular axes
resultantthe single vector obtained by adding two or more vectors together
projectilean object in flight after being thrown or launched, moving under gravity alone
time of flightthe total time a projectile remains airborne, from launch to landing
rangethe horizontal distance a projectile travels from launch to landing
centripetal accelerationthe acceleration of an object moving in a circular path, always directed toward the centre, with magnitude v²/R
angular speedthe rate at which an object's angular position around a circle changes with time, symbol ω
🔒

Model exam answers, grammar & audio

You have read the summary. The board-ready model answers, grammar notes, one-touch audio and writing practice for this chapter are part of Lipi©.

Unlock free with any language course

See it, understand it, hear it read aloud, then write the exam answer with confidence, for a fraction of a tutor cost.