A cricketer can throw a ball to a maximum horizontal distance of $100\ \text{m}$. How much high above the ground can the cricketer throw the same ball?

> [!TLDR] > The maximum horizontal range of a projectile is achieved when launched at $45^\circ$, and its maximum vertical height is achieved when launched at $90^\circ$ (vertically upwards). Since the initial speed of the ball is the same in both cases, we can relate the given maximum range to the maximum possible height, finding it to be $\boxed{50\ \text{m}}$. When a cricketer throws a ball, the initial speed ($u$) with which the ball leaves their hand is determined by their effort. This initial speed is a fundamental quantity that dictates how far or how high the ball can go. The trajectory of the ball, whether it's a long throw or a high throw, is governed by the principles of projectile motion, where the only significant force acting on the ball after it leaves the hand is gravity (neglecting air resistance). The problem presents two scenarios, both involving the *same* ball, implying the *same initial speed* $u$. 1. **Maximum Horizontal Distance (Range):** To achieve the maximum horizontal distance, the ball must be thrown at a specific angle. 2. **Maximum Vertical Height:** To achieve the maximum possible height, the ball must be thrown straight upwards. Our strategy will be to first use the given maximum horizontal range to determine the value of $u^2/g$, and then use this value to calculate the maximum vertical height. 1. **Identify the given information and the constant quantity:** The maximum horizontal distance (range) is $R_{max} = 100\ \text{m}$. The initial speed $u$ of the ball is the same for both the maximum range throw and the maximum height throw. The acceleration due to gravity is $g$. 2. **Recall the formula for the horizontal range of a projectile:** The horizontal range $R$ of a projectile launched with initial speed $u$ at an angle $\theta$ with the horizontal is given by: $$R = \frac{u^2 \sin(2\theta)}{g}$$ 3. **Determine the condition for maximum horizontal range:** For $R$ to be maximum, the term $\sin(2\th

A cricketer can throw a ball to a maximum horizontal distance of 100m. How much high above the ground can the

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100\ \text{m}

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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 maximum horizontal range of a projectile is achieved when launched at $45^\circ$ , and its maximum vertical height is achieved when launched at $90^\circ$ (vertically upwards). Since the initial speed of the ball is the same in both cases, we can relate the given maximum range to the maximum possible height, finding it to be $\boxed{50\ \text{m}}$ .

When a cricketer throws a ball, the initial speed ( $u$ ) with which the ball leaves their hand is determined by their effort. This initial speed is a fundamental quantity that dictates how far or how high the ball can go. The trajectory of the ball, whether it's a long throw or a high throw, is governed by the principles of projectile motion, where the only significant force acting on the ball after it leaves the hand is gravity (neglecting air resistance).

The problem presents two scenarios, both involving the same ball, implying the same initial speed $u$ .

Maximum Horizontal Distance (Range): To achieve the maximum horizontal distance, the ball must be thrown at a specific angle.
Maximum Vertical Height: To achieve the maximum possible height, the ball must be thrown straight upwards.

Our strategy will be to first use the given maximum horizontal range to determine the value of $u^2/g$ , and then use this value to calculate the maximum vertical height.

Identify the given information and the constant quantity:

The maximum horizontal distance (range) is $R_{max} = 100\ \text{m}$ .

The initial speed $u$ of the ball is the same for both the maximum range throw and the maximum height throw. The acceleration due to gravity is $g$ .
Recall the formula for the horizontal range of a projectile:

The horizontal range $R$ of a projectile launched with initial speed $u$ at an angle $\theta$ with the horizontal is given by:

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

Determine the condition for maximum horizontal range: For $R$ to be maximum, the term $\sin(2\theta)$ must be maximum. The maximum value of $\sin(2\theta)$ is $1$ , which occurs when $2\theta = 90^\circ$ , or $\theta = 45^\circ$ . Therefore, the maximum horizontal range $R_{max}$ is:

R_{max} = \frac{u^2 (1)}{g} = \frac{u^2}{g}

Use the given $R_{max}$ to find $u^2/g$ : We are given $R_{max} = 100\ \text{m}$ . Substituting this into the formula from Step 3:

100\ \text{m} = \frac{u^2}{g}

This gives us a crucial relationship: $u^2/g = 100\ \text{m}$. We will use this in the next part of the problem.

5. Determine the condition for maximum vertical height:

To throw the ball as high as possible, the cricketer must throw it straight upwards. This corresponds to a launch angle of $\theta = 90^\circ$ with the horizontal. In this case, all the initial velocity is directed vertically.

Recall the formula for the maximum height of a projectile (or use kinematics for vertical motion): …

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Concept: Prove the Maximizing Angle by Calculus, Then Switch to Energy Conservation for the Height

Method: Derivative Test for $\theta_{\max}$ (Not "Inspection of $\sin$ "), Then Work-Energy Theorem for $H_{\max}$ (Not the Kinematic Height Formula)

The existing solutions both quote " $R$ is maximum when $\sin2\theta=1$ " as a known fact, then use the kinematic formula $H=\dfrac{u^2\sin^2\theta}{2g}$ for the height. This method proves the maximizing angle from calculus instead of quoting the sine function's known maximum, and finds the vertical throw's height from energy conservation instead of a kinematics formula — genuinely different tools for both halves.

Part 1 — Prove $\theta=45^\circ$ maximizes range, by differentiating

The range as a function of launch angle is $R(\theta) = \dfrac{u^2\sin2\theta}{g}$ . To find where this is maximized, set its derivative with respect to $\theta$ to zero:

\frac{dR}{d\theta} = \frac{u^2}{g}\cdot 2\cos2\theta = 0 \quad\Longrightarrow\quad \cos2\theta = 0 \quad\Longrightarrow\quad 2\theta = 90^\circ \quad\Longrightarrow\quad \theta = 45^\circ

Checking this is a maximum (not a minimum): the second derivative $\dfrac{d^2R}{d\theta^2} = -\dfrac{4u^2}{g}\sin2\theta$ is negative at $\theta=45^\circ$ (since $\sin90^\circ=1>0$ ), confirming a maximum. So:

R_{\max} = R(45^\circ) = \frac{u^2\sin90^\circ}{g} = \frac{u^2}{g}

Part 2 — Use the given $R_{\max}$ to pin down $u^2/g$

100\ \text{m} = \frac{u^2}{g} \quad\Longrightarrow\quad u^2 = 100g

This is the same ball, so this $u^2$ is what carries over to the vertical throw.

Part 3 — Vertical-throw height, via the work-energy theorem (not a kinematic formula)

When thrown straight up, all of the ball's initial kinetic energy converts to gravitational potential energy at the top (where it momentarily stops, $v=0$ ):

\underbrace{\tfrac12 m u^2}_{\text{initial KE}} = \underbrace{mgH_{\max}}_{\text{PE gained}}

The mass $m$ cancels from both sides — worth noting, since it means the answer is independent of how heavy the ball is: …

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Q.A cricketer can throw a ball to a maximum horizontal distance of 100 m100\ \text{m}100 m. How much high above the ground can the cricketer throw the same ball?

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Q.A cricketer can throw a ball to a maximum horizontal distance of $100\ \text{m}$ . How much high above the ground can the cricketer throw the same ball?

Unlock everything free for 14 days