# Projectile Motion: The Complete Guide for IOE Entrance

This guide provides a structured and in-depth understanding of projectile motion, covering all the essential concepts, formulas, and problem-solving techniques required for the IOE entrance exam.

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### 1. What is Projectile Motion?

[cite_start]A **projectile** is any object that is thrown or projected into the air and then moves under the influence of gravity alone[cite: 2]. We make a few key assumptions for simplifying our calculations:
* [cite_start]Acceleration due to gravity (`g`) is constant (both in magnitude and direction)[cite: 7].
* Air resistance is negligible.

[cite_start]The path followed by a projectile is called its **trajectory**, which is always a parabola[cite: 15, 27].

[cite_start]**The Golden Rule:** Analyze horizontal and vertical motion **independently**[cite: 6].
* [cite_start]**Horizontal Motion (x-axis):** No acceleration ($a_x = 0$)[cite: 9]. [cite_start]The horizontal velocity is always **constant**[cite: 11, 12].
* [cite_start]**Vertical Motion (y-axis):** Constant downward acceleration due to gravity ($a_y = -g$)[cite: 7].

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### 2. Ground-to-Ground Projection

[cite_start]This is the most common scenario, where a projectile is launched from the ground and lands back on the ground[cite: 16].

Let's say a projectile is launched with an initial velocity **u** at an angle **θ** with the horizontal.

* [cite_start]**Initial Horizontal Velocity ($u_x$):** $u_x = u \cos\theta$ [cite: 19]
* [cite_start]**Initial Vertical Velocity ($u_y$):** $u_y = u \sin\theta$ [cite: 19]

#### a) Equation of Trajectory

[cite_start]This equation describes the parabolic path of the projectile[cite: 15].

We know:
1.  [cite_start]Horizontal distance: $x = u_x t = (u \cos\theta) t \implies t = \frac{x}{u \cos\theta}$ [cite: 23]
2.  [cite_start]Vertical distance: $y = u_y t - \frac{1}{2}gt^2$ [cite: 25]

Substitute the value of `t` from (1) into (2):
[cite_start]$y = (u \sin\theta) \left(\frac{x}{u \cos\theta}\right) - \frac{1}{2}g\left(\frac{x}{u \cos\theta}\right)^2$ [cite: 31]

This gives us the two standard forms of the trajectory equation:

> [cite_start]**Equation 1:** $y = x \tan\theta - \frac{g x^2}{2u^2 \cos^2\theta}$ [cite: 27, 32]
>
> [cite_start]**Equation 2 (in terms of Range, R):** $y = x \tan\theta \left(1 - \frac{x}{R}\right)$ [cite: 27, 416]

#### b) Time of Flight (T)

[cite_start]This is the total time the projectile remains in the air[cite: 33]. [cite_start]The net vertical displacement for a ground-to-ground projection is zero[cite: 37, 38].
Using $s_y = u_y t - \frac{1}{2}gt^2$:
[cite_start]$0 = (u \sin\theta) T - \frac{1}{2}gT^2$ [cite: 34]
$T \left(u \sin\theta - \frac{1}{2}gT\right) = 0$
Since $T \neq 0$, we get:

> [cite_start]**Time of Flight (T):** $T = \frac{2u \sin\theta}{g} = \frac{2u_y}{g}$ [cite: 35]

* **Time to reach maximum height ($t_a$):** This is half of the total time of flight. $t_a = \frac{T}{2} = \frac{u\sin\theta}{g}$

#### c) Maximum Height (H)

[cite_start]This is the highest vertical distance reached by the projectile[cite: 83]. [cite_start]At this point, the final vertical velocity ($v_y$) is zero[cite: 86].
Using $v_y^2 = u_y^2 + 2a_y s_y$:
[cite_start]$0^2 = (u \sin\theta)^2 + 2(-g)H$ [cite: 89, 90, 92]

> [cite_start]**Maximum Height (H):** $H = \frac{u^2 \sin^2\theta}{2g} = \frac{u_y^2}{2g}$ [cite: 95]

#### d) Horizontal Range (R)

[cite_start]This is the total horizontal distance covered during the time of flight[cite: 52].
Using $s_x = u_x t$:
[cite_start]$R = (u \cos\theta) T = (u \cos\theta) \left(\frac{2u \sin\theta}{g}\right)$ [cite: 53, 54]

> [cite_start]**Horizontal Range (R):** $R = \frac{u^2 (2\sin\theta\cos\theta)}{g} = \frac{u^2 \sin(2\theta)}{g} = \frac{2u_x u_y}{g}$ [cite: 55]

* [cite_start]**Maximum Range:** The range is maximum when $\sin(2\theta) = 1$ [cite: 60][cite_start], which means $2\theta = 90^\circ$, so $\theta = 45^\circ$[cite: 56, 61].
    > [cite_start]$R_{max} = \frac{u^2}{g}$ (for $\theta = 45^\circ$) [cite: 57]
* [cite_start]**Complementary Angles:** The range is the same for two angles of projection, $\theta$ and $(90^\circ - \theta)$, for the same initial speed `u`[cite: 64, 65, 67].

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### 3. Important Relations & Special Cases

#### a) Relation between R and H

[cite_start]A very useful formula for quick calculations[cite: 162]:
$R = \frac{u^2 (2\sin\theta\cos\theta)}{g}$ and $H = \frac{u^2 \sin^2\theta}{2g}$
Dividing R by H and rearranging gives:

> [cite_start]$R = 4H \cot\theta$ [cite: 161, 166]

**Special Case:** If Range equals Maximum Height ($R=H$):
[cite_start]$H = 4H \cot\theta \implies \cot\theta = \frac{1}{4} \implies \tan\theta = 4$ [cite: 168, 170, 171]
[cite_start]$\theta = \tan^{-1}(4) \approx 76^\circ$ [cite: 172]

#### b) Velocity at Any Time `t`

* [cite_start]**Horizontal velocity ($v_x$):** $v_x = u \cos\theta$ (always constant) [cite: 108]
* [cite_start]**Vertical velocity ($v_y$):** $v_y = u \sin\theta - gt$ [cite: 109]
* [cite_start]**Resultant velocity ($\vec{v}$):** $\vec{v} = v_x \hat{i} + v_y \hat{j} = (u \cos\theta)\hat{i} + (u \sin\theta - gt)\hat{j}$ [cite: 112, 113]
* **Speed:** $|\vec{v}| = \sqrt{v_x^2 + v_y^2}$
* [cite_start]**Angle with horizontal ($\alpha$):** $\tan\alpha = \frac{v_y}{v_x} = \frac{u\sin\theta - gt}{u\cos\theta}$ [cite: 100, 102]

#### c) When is velocity perpendicular to initial velocity?

[cite_start]The condition is $\vec{v} \cdot \vec{u} = 0$[cite: 123].
[cite_start]$\big((u \cos\theta)\hat{i} + (u \sin\theta - gt)\hat{j}\big) \cdot \big((u \cos\theta)\hat{i} + (u \sin\theta)\hat{j}\big) = 0$ [cite: 122, 123]
[cite_start]$u^2\cos^2\theta + u^2\sin^2\theta - (gt)(u\sin\theta) = 0$ [cite: 124]
[cite_start]$u^2 = gut\sin\theta$ [cite: 125]

> [cite_start]**Time when $\vec{v} \perp \vec{u}$:** $t = \frac{u}{g \sin\theta}$ [cite: 116, 126]

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### 4. Projection from a Height (H)

#### Case 1: Projected Horizontally

[cite_start]A body is thrown horizontally with speed `u` from a height `H`[cite: 311].
* Initial horizontal velocity: $u_x = u$
* [cite_start]Initial vertical velocity: $u_y = 0$ [cite: 325]

* **Time to reach ground (T):**
    [cite_start]Use $s_y = u_y t + \frac{1}{2}a_y t^2$[cite: 312]. Taking downward as positive displacement (`H`) and downward acceleration (`g`):
    [cite_start]$H = (0)T + \frac{1}{2}gT^2 \implies T = \sqrt{\frac{2H}{g}}$ [cite: 314]
* **Horizontal Range (R):**
    [cite_start]$R = u_x T \implies R = u \sqrt{\frac{2H}{g}}$ [cite: 333, 334]
* **Velocity on striking the ground:**
    $v_x = u$
    [cite_start]$v_y = gt$ [cite: 339]
    Resultant Speed: $|\vec{v}| [cite_start]= \sqrt{v_x^2 + v_y^2} = \sqrt{u^2 + (gT)^2} = \sqrt{u^2 + 2gH}$ [cite: 343, 346]
    [cite_start]Angle of strike ($\alpha$): $\tan\alpha = \frac{v_y}{v_x} = \frac{gT}{u} = \frac{\sqrt{2gH}}{u}$ [cite: 340]

#### Case 2: Projected at an Angle $\theta$

* **Angle of Elevation (Upwards):**
    [cite_start]$u_x = u\cos\theta$, $u_y = u\sin\theta$[cite: 285].
    [cite_start]Use $s_y = u_y t + \frac{1}{2}a_y t^2$[cite: 281]. The net vertical displacement is `-H`.
    [cite_start]$-H = (u\sin\theta)t - \frac{1}{2}gt^2$[cite: 282]. This is a quadratic equation in `t` which can be solved.
* **Angle of Depression (Downwards):**
    [cite_start]$u_x = u\cos\theta$, $u_y = -u\sin\theta$[cite: 306, 307, 321].
    [cite_start]$-H = (-u\sin\theta)t - \frac{1}{2}gt^2 \implies H = (u\sin\theta)t + \frac{1}{2}gt^2$[cite: 308, 309].

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### 5. Energy and Momentum

#### At the Highest Point (H):
* [cite_start]Velocity is purely horizontal: $v = u\cos\theta$[cite: 84].
* [cite_start]**Kinetic Energy ($KE_H$):** $KE_H = \frac{1}{2}m(u\cos\theta)^2 = (\frac{1}{2}mu^2)\cos^2\theta = KE_{initial} \cos^2\theta$[cite: 433].
* [cite_start]**Potential Energy ($PE_H$):** $PE_H = mgH = mg\left(\frac{u^2\sin^2\theta}{2g}\right) = (\frac{1}{2}mu^2)\sin^2\theta = KE_{initial} \sin^2\theta$[cite: 445, 446, 447].
* **Change in Momentum (from launch to H):**
    [cite_start]$\Delta\vec{P} = \vec{P}_f - \vec{P}_i = (mu\cos\theta)\hat{i} - (mu\cos\theta\hat{i} + mu\sin\theta\hat{j}) = -mu\sin\theta\hat{j}$[cite: 458, 462, 463].
    Magnitude: $|\Delta\vec{P}| [cite_start]= mu\sin\theta$[cite: 464].

#### For the Entire Flight (Ground to Ground):
* [cite_start]**Change in Kinetic Energy:** $\Delta KE = 0$ (since speed is the same)[cite: 532].
* [cite_start]**Change in Potential Energy:** $\Delta PE = 0$[cite: 532].
* **Change in Momentum:**
    [cite_start]$\vec{v}_f = u\cos\theta\hat{i} - u\sin\theta\hat{j}$ [cite: 513]
    [cite_start]$\vec{v}_i = u\cos\theta\hat{i} + u\sin\theta\hat{j}$ [cite: 430]
    [cite_start]$\Delta\vec{P} = m(\vec{v}_f - \vec{v}_i) = m(-2u\sin\theta\hat{j})$[cite: 529, 530, 533].
    Magnitude: $|\Delta\vec{P}| [cite_start]= 2mu\sin\theta$[cite: 531, 534].

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### 6. Example Problems

**Example 1:** A cricket ball is hit at $45^\circ$ to the horizontal with a kinetic energy of K. What is the kinetic energy at the highest point?

**Solution:**
[cite_start]At the highest point, $v = u\cos\theta$[cite: 431].
[cite_start]$KE_{highest} = \frac{1}{2}m(u\cos\theta)^2 = (\frac{1}{2}mu^2)\cos^2\theta = K \cos^2\theta$[cite: 433].
Given $\theta = 45^\circ$, $\cos(45^\circ) = \frac{1}{\sqrt{2}}$.
$KE_{highest} = K \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{K}{2}$.

**Example 2:** A body is projected with velocity $\vec{u} = (3\hat{i} + 4\hat{j})$ m/s. Find the horizontal range. [cite_start](Assume $g=10$ m/s²) [cite: 262]

**Solution:**
From the vector, we have:
[cite_start]$u_x = 3$ m/s [cite: 263]
[cite_start]$u_y = 4$ m/s [cite: 263]
[cite_start]Using the formula $R = \frac{2u_x u_y}{g}$[cite: 270]:
[cite_start]$R = \frac{2 \times 3 \times 4}{10} = \frac{24}{10} = 2.4$ m[cite: 270].

[cite_start]**Example 3:** For a projectile, the range R is equal to the maximum height H. Find the angle of projection[cite: 168].

**Solution:**
[cite_start]We use the relation $R = 4H\cot\theta$[cite: 161].
[cite_start]Given $R=H$[cite: 168]:
[cite_start]$H = 4H\cot\theta$ [cite: 170]
$1 = 4\cot\theta$
[cite_start]$\tan\theta = 4$ [cite: 171]
[cite_start]$\theta = \tan^{-1}(4)$[cite: 172].