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Q.What is projectile ? Give two examples.

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2019Subjective· 2mImportance★★★★★
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Concept understanding — Projectile Motion

Projectile Motion — From Intuition to Precision

Imagine you throw a ball to a friend. It doesn't travel in a straight line — it rises, slows down, then curves downward and falls. That curved path is a projectile's trajectory. The ball is a projectile: any object that is launched into the air and then moves only under the influence of gravity (and air resistance, which we ignore for now).

The key intuition: once the ball leaves your hand, the only force acting on it is gravity pulling it straight down. There is no forward force after release. The ball keeps moving forward because of inertia — it wants to keep going in a straight line at constant speed. But gravity keeps pulling it down, so the forward motion and downward acceleration combine to produce a curved path.


The Precise Statement

Projectile motion is the two-dimensional motion of an object launched into the air, subject only to the constant downward acceleration due to gravity (g≈9.8 m/s2g \approx 9.8 \, \text{m/s}^2). Air resistance is neglected.

We break the motion into two independent components:

  • Horizontal motion: No acceleration (ax=0a_x = 0). So horizontal velocity vxv_x is constant.
  • Vertical motion: Constant downward acceleration (ay=−ga_y = -g). So vertical velocity vyv_y changes linearly with time.

The independence of these components is the central idea — what happens vertically does not affect what happens horizontally, and vice versa.


The Equations (for a projectile launched with initial speed uu at angle θ\theta above horizontal)

First, resolve the initial velocity:

ux=ucos⁡θ,uy=usin⁡θu_x = u \cos\theta, \quad u_y = u \sin\theta

Horizontal motion (constant velocity):

x=uxt=(ucos⁡θ)tx = u_x t = (u \cos\theta) t

Vertical motion (constant acceleration −g-g):

vy=uy−gt=usin⁡θ−gtv_y = u_y - gt = u \sin\theta - gt

y=uyt−12gt2=(usin⁡θ)t−12gt2y = u_y t - \frac{1}{2} g t^2 = (u \sin\theta) t - \frac{1}{2} g t^2


Key Results You Must Know

Important

Time of flight TT: total time the projectile stays in the air (until y=0y=0 again).

T=2usin⁡θgT = \frac{2u \sin\theta}{g}

Important

Maximum height HH: the highest vertical position reached (when vy=0v_y = 0).

H=u2sin⁡2θ2gH = \frac{u^2 \sin^2\theta}{2g}

Important

Range RR: the horizontal distance covered when it returns to launch height.

R=u2sin⁡2θgR = \frac{u^2 \sin 2\theta}{g}

Note

The range is maximum when sin⁡2θ=1\sin 2\theta = 1, i.e., θ=45∘\theta = 45^\circ. For a given speed, 45∘45^\circ gives the farthest throw.


The Trajectory Equation (Path Shape)

Eliminate tt from the xx and yy equations to get yy as a function of xx:

y=xtan⁡θ−gx22u2cos⁡2θy = x \tan\theta - \frac{g x^2}{2 u^2 \cos^2\theta}

This is a parabola — the signature shape of projectile motion.


Common Mistake to Avoid …

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