A fighter plane is flying horizontally at an altitude of 1.5 km with speed 720 km/h. At what angle of sight (w

NCERT Exemplar · Q24

Q. A fighter plane is flying horizontally at an altitude of 1.5 km with speed 720 km/h. At what angle of sight (w.r.t. horizontal) when the target is seen, should the pilot drop the bomb in order to attack the target?

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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

g \approx 9.8 \, \text{m/s}^2

We break the motion into two independent components:

Horizontal motion: No acceleration ( $a_x = 0$ ). So horizontal velocity $v_x$ is constant.
Vertical motion: Constant downward acceleration ( $a_y = -g$ ). So vertical velocity $v_y$ 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 $u$ at angle $\theta$ above horizontal)

First, resolve the initial velocity:

u_x = u \cos\theta, \quad u_y = u \sin\theta

Horizontal motion (constant velocity):

x = u_x t = (u \cos\theta) t

Vertical motion (constant acceleration $-g$ ):

v_y = u_y - gt = u \sin\theta - gt

y = 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 $T$ : total time the projectile stays in the air (until $y=0$ again).

$T = \frac{2u \sin\theta}{g}$

Important

Maximum height $H$ : the highest vertical position reached (when $v_y = 0$ ).

$H = \frac{u^2 \sin^2\theta}{2g}$

Important

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

$R = \frac{u^2 \sin 2\theta}{g}$

Note

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

The Trajectory Equation (Path Shape)

Eliminate $t$ from the $x$ and $y$ equations to get $y$ as a function of $x$ :

y = 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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The bomb must be released while the target is sighted at about 23.2° below the horizontal, so that the bomb's forward drift during its fall exactly covers the remaining horizontal distance to the target.

Setting up the problem

Once released, the bomb keeps moving forward with the plane's horizontal speed while falling under gravity — independent horizontal and vertical motions, just like any projectile. The pilot must release the bomb while it is still short of the target, so that by the time it falls to the ground it has drifted forward exactly onto the target. The "angle of sight" is the angle below the horizontal at which the target is seen at that release instant.

Step 1 — Convert to SI units

h = 1.5\ \text{km} = 1500\ \text{m}, \qquad v_x = 720\ \text{km/h} \times \frac{5}{18} = 200\ \text{m/s}

Step 2 — Time to fall from 1500 m

The bomb starts with zero vertical velocity, so using $h = \frac{1}{2}gt^2$ with $g = 9.8\ \text{m/s}^2$ :

1500 = \frac{1}{2}(9.8)t^2 \implies t^2 = \frac{1500}{4.9} \approx 306.1 \implies t \approx 17.5\ \text{s}

Step 3 — Horizontal distance covered while falling …

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Q.A fighter plane is flying horizontally at an altitude of 1.5 km with speed 720 km/h. At what angle of sight (w.r.t. horizontal) when the target is seen, should the pilot drop the bomb in order to attack the target?

Concept understanding — Projectile Motion

Projectile Motion — From Intuition to Precision

The Precise Statement

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

Key Results You Must Know

The Trajectory Equation (Path Shape)

Common Mistake to Avoid …

Setting up the problem

Step 1 — Convert to SI units

Step 2 — Time to fall from 1500 m

Step 3 — Horizontal distance covered while falling …

Concept: Derive One General Formula Before Plugging In Any Numbers

Steps

Q.A fighter plane is flying horizontally at an altitude of 1.5 km with speed 720 km/h. At what angle of sight (w.r.t. horizontal) when the target is seen, should the pilot drop the bomb in order to attack the target?

Setting up the problem

Step 1 — Convert to SI units

Step 2 — Time to fall from 1500 m

Step 3 — Horizontal distance covered while falling …

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