A ball is thrown from a roof top at an angle of $45^\circ$ above the horizontal. It hits the ground a few seconds later. At what point during its motion, does the ball have (a) greatest speed. (b) smallest speed. (c) greatest acceleration? Explain

A ball is thrown from a roof top at an angle of 45^° above the horizontal. It hits the ground a few seconds la

NCERT Exemplar · Q18

45^\circ

(a) greatest speed.

(b) smallest speed.

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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 ball's horizontal velocity is constant, while its vertical velocity changes due to gravity.

Greatest speed occurs just before hitting the ground, as the ball has gained maximum downward vertical velocity.
Smallest speed occurs at the highest point of its trajectory, where its vertical velocity momentarily becomes zero.
The acceleration is constant throughout the motion, always equal to the acceleration due to gravity, $g$ .

When a ball is thrown into the air, its motion is governed by the force of gravity. According to Newton's Second Law, force causes acceleration. In this scenario, the only significant force acting on the ball (ignoring air resistance) is gravity, which acts vertically downwards. This means the ball experiences a constant acceleration in the vertical direction, while its horizontal motion remains unaffected. Understanding how these horizontal and vertical components of velocity change allows us to determine the points of greatest and smallest speed, and the nature of its acceleration.

Let's break down each part of the question:

(a) Greatest Speed

Understanding Speed: Speed is the magnitude of the velocity vector. If the velocity components are $v_x$ (horizontal) and $v_y$ (vertical), the speed $v$ is given by $v = \sqrt{v_x^2 + v_y^2}$ .
Horizontal Velocity Component ( $v_x$ ): Since there is no horizontal force (ignoring air resistance), the horizontal acceleration is zero. This means the horizontal component of velocity, $v_x$ , remains constant throughout the ball's flight.
Vertical Velocity Component ( $v_y$ ): The vertical component of velocity, $v_y$ , is continuously affected by gravity. It decreases as the ball moves upwards, becomes zero at the highest point of its trajectory, and then increases in magnitude (downwards) as the ball falls.
Maximizing Speed: To maximize the overall speed $v = \sqrt{v_x^2 + v_y^2}$ , we need to maximize the magnitude of the vertical velocity component, $|v_y|$ , because $v_x$ is constant.
Point of Greatest Vertical Velocity: The ball is thrown from a rooftop, meaning it can fall a greater vertical distance than it initially rose. As the ball falls from its highest point to the ground, gravity continuously accelerates it downwards, causing its downward vertical speed to increase. The magnitude of the vertical velocity will be greatest just before the ball hits the ground.
Conclusion: Since $v_x$ is constant and $|v_y|$ is greatest just before hitting the ground, the overall speed $v$ will be greatest at this point.

(b) Smallest Speed

Minimizing Speed: To minimize the overall speed $v = \sqrt{v_x^2 + v_y^2}$ , we need to minimize the magnitude of the vertical velocity component, $|v_y|$ , because $v_x$ is constant and never zero (unless the ball is thrown straight up, which is not the case here as it's thrown at $45^\circ$ ).
Point of Zero Vertical Velocity: As the ball travels upwards, its vertical velocity decreases due to gravity. At the very peak of its trajectory, the ball momentarily stops moving vertically before it starts to fall downwards. At this instant, its vertical velocity $v_y$ is zero.
Speed at the Peak: At the highest point, $v_y = 0$ . The speed of the ball is then $v = \sqrt{v_x^2 + 0^2} = v_x$ . Since $v_x$ is constant and non-zero, this represents the minimum possible speed for the ball during its flight.
Conclusion: The smallest speed occurs at the highest point of the ball's trajectory. …

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Concept: Treat Speed-Squared as an Explicit Function of Time and Find Its Global Extremum by Calculus

Method: Extremum Analysis of $v^2(t) = v_x^2+v_y(t)^2$ as a Quadratic in $t$

Rather than reasoning qualitatively about "gravity speeds the ball up as it falls," this method writes the squared speed as an explicit function of time -- a quadratic, since $v_y(t)$ is linear in $t$ -- and locates its single critical point using ordinary calculus, then argues about the two domain endpoints to settle where the maximum occurs.

Step 1 -- Write $v^2(t)$ explicitly

Take the roof as the origin, launch angle $45^\circ$ , launch speed $u$ , so $v_x = u\cos45^\circ$ (constant for the whole flight) and $v_y(t) = u\sin45^\circ - gt$ (positive while rising, zero at the peak, negative and growing in magnitude while falling). The squared speed is:

v^2(t) = v_x^2 + v_y(t)^2 = v_x^2 + \big(u\sin45^\circ - gt\big)^2

This is a quadratic in $t$ , opening upward (the coefficient of $t^2$ is $g^2>0$ ).

Step 2 -- Find the critical point by differentiating

\frac{d}{dt}\big[v^2(t)\big] = 2\big(u\sin45^\circ-gt\big)(-g) = -2g\,v_y(t)

Setting this to zero: $v_y(t)=0$ , which happens at exactly one instant, $t^\ast = \dfrac{u\sin45^\circ}{g}$ -- this is precisely the time the ball reaches its highest point.

Step 3 -- Classify this critical point

Since $v^2(t)$ is an upward-opening parabola in $t$ (coefficient of $t^2$ is $+g^2$ ), its unique critical point $t^\ast$ must be a global minimum over all $t$ -- an upward parabola has no interior maximum, only a single minimum at its vertex. So speed is smallest exactly at $t=t^\ast$ , the highest point, and nowhere else. This settles part (b) directly from the shape of the function, without comparing candidate points one by one.

Step 4 -- Since there is no interior maximum, the greatest speed must occur at a domain endpoint

An upward parabola with no interior maximum can only be largest at the edges of whatever time interval is being considered -- here, $t=0$ (the instant of launch) or $t=T$ (the instant of landing). Compare the two:

At $t=0$ : $v^2(0) = v_x^2 + (u\sin45^\circ)^2 = u^2$ (the launch speed squared).
At $t=T$ (landing): since the ball is thrown from a rooftop and lands at ground level, it falls through a greater height than it rose through during the ascent (the ball descends past its own launch height and keeps going). This means $|v_y(T)| > |v_y(0)| = u\sin45^\circ$ (a longer, purely-gravity-driven fall past the launch point adds more speed than the initial upward throw removed), so $v^2(T) = v_x^2+v_y(T)^2 > v_x^2+v_y(0)^2 = v^2(0)$ .

Since the parabola has no interior maximum and the landing endpoint gives a strictly larger value than the launch endpoint, $t=T$ (landing, just before impact) is where $v^2(t)$ -- and hence speed -- is greatest.

Step 5 -- Acceleration needs no extremum analysis at all …

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Q.A ball is thrown from a roof top at an angle of 45∘45^\circ45∘ above the horizontal. It hits the ground a few seconds later. At what point during its motion, does the ball have

(a) Greatest Speed

(b) Smallest Speed

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Q.A ball is thrown from a roof top at an angle of $45^\circ$ above the horizontal. It hits the ground a few seconds later. At what point during its motion, does the ball have

Unlock everything free for 14 days