Projectile Motion — Revision Notes

⚡ 30-Second Revision

Horizontal Motion: — $v_x = u cos\theta$ (constant), $a_x = 0$, $x = (u cos\theta)t$
Vertical Motion: — $u_y = u sin\theta$, $a_y = -g$, $v_y = u sin\theta - gt$, $y = (u sin\theta)t - \frac{1}{2}gt^2$
Time of Flight (T): — $T = \frac{2u sin\theta}{g}$
Maximum Height (H): — $H = \frac{u^2 sin^2\theta}{2g}$
Horizontal Range (R): — $R = \frac{u^2 sin 2\theta}{g}$
Max Range Angle: — $heta = 45^circ$, $R_{max} = \frac{u^2}{g}$
Complementary Angles: — $heta$ and $(90^circ - \theta)$ give same $R$.
Velocity at Max Height: — $v = u cos\theta$ (vertical component is zero).
Trajectory Equation: — $y = x \tan\theta - \frac{gx^2}{2u^2 cos^2\theta}$ (parabolic)

2-Minute Revision

Projectile motion is the 2D movement of an object under gravity, neglecting air resistance. The key is to analyze horizontal and vertical motions independently. Horizontally, velocity ($u cos\theta$) is constant, as acceleration is zero. Vertically, the motion is uniformly accelerated by gravity ($g$ downwards), so velocity changes. The initial vertical velocity is $u sin\theta$. The path is a parabola.

Key formulas to remember:

1
Time of Flight ($T$): — Total time in air, $T = \frac{2u sin\theta}{g}$.
2
Maximum Height ($H$): — Highest point reached, $H = \frac{u^2 sin^2\theta}{2g}$. At this point, vertical velocity is zero, but horizontal velocity ($u cos\theta$) is still present.
3
Horizontal Range ($R$): — Total horizontal distance, $R = \frac{u^2 sin 2\theta}{g}$. Maximum range occurs at $heta = 45^circ$, where $R_{max} = u^2/g$. Complementary angles (e.g., $30^circ$ and $60^circ$) yield the same range.

For horizontal projection from a height, $u_y = 0$, and time of flight is found from $h = \frac{1}{2}gt^2$. Relative motion between two projectiles under gravity implies zero relative acceleration, so their relative velocity is constant. Practice applying these formulas and concepts to various problem types.

5-Minute Revision

Projectile motion is the study of objects moving in two dimensions under the sole influence of gravity, with air resistance typically ignored. The core principle is the independence of horizontal and vertical motion. This means we can analyze each component separately.

Horizontal Motion:

No horizontal forces, so horizontal acceleration $a_x = 0$.
Horizontal velocity $v_x = u_x = u cos\theta$ remains constant throughout the flight.
Horizontal distance $x = u_x t = (u cos\theta)t$.

Vertical Motion:

Constant downward acceleration $a_y = -g$.
Initial vertical velocity $u_y = u sin\theta$.
Vertical velocity at time $t$: $v_y = u sin\theta - gt$.
Vertical displacement at time $t$: $y = (u sin\theta)t - \frac{1}{2}gt^2$.
At maximum height, $v_y = 0$.

Key Parameters & Formulas:

1
Time of Flight ($T$): — The total time the projectile is in the air. For ground-to-ground projection, $y=0$ at $t=T$. $T = \frac{2u sin\theta}{g}$.
2
Maximum Height ($H$): — The highest vertical point reached. Occurs when $v_y = 0$. $H = \frac{u^2 sin^2\theta}{2g}$.
3
Horizontal Range ($R$): — The total horizontal distance covered. R = u_x T = (u cos\theta) left(\frac{2u sin\theta}{g}\right) = \frac{u^2 sin 2\theta}{g}.

Important Points:

Maximum Range: — Achieved at $heta = 45^circ$, $R_{max} = u^2/g$.
Complementary Angles: — Angles $heta$ and $(90^circ - \theta)$ give the same range for a given initial speed.
Velocity at Max Height: — Only the horizontal component $u cos\theta$ is present; $v_y = 0$.
Trajectory: — The path is a parabola, described by $y = x \tan\theta - \frac{gx^2}{2u^2 cos^2\theta}$.

Worked Mini-Example: A ball is thrown with $u = 10sqrt{2},\text{m/s}$ at $heta = 45^circ$. Find its range ($g=10,\text{m/s}^2$). $R = \frac{u^2 sin 2\theta}{g} = \frac{(10sqrt{2})^2 sin(2 \times 45^circ)}{10} = \frac{100 \times 2 \times sin 90^circ}{10} = \frac{200 \times 1}{10} = 20,\text{m}$.

Remember to apply these concepts carefully, especially when dealing with projection from a height or relative motion problems. Always resolve initial velocities into components.

Prelims Revision Notes

Projectile motion is a critical topic for NEET, focusing on 2D motion under gravity. The core idea is the independence of horizontal and vertical motion. This means you can analyze them separately.

1. Initial Velocity Components:

* If initial velocity is $u$ at angle $heta$ with horizontal: * Horizontal component: $u_x = u cos\theta$ * Vertical component: $u_y = u sin\theta$

2. Motion Equations (Kinematics):

* Horizontal (constant velocity): * Acceleration $a_x = 0$ * Velocity $v_x = u_x = u cos\theta$ (constant) * Displacement $x = u_x t = (u cos\theta)t$ * Vertical (uniformly accelerated): * Acceleration $a_y = -g$ (downwards) * Velocity $v_y = u_y + a_y t = u sin\theta - gt$ * Displacement $y = u_y t + \frac{1}{2}a_y t^2 = (u sin\theta)t - \frac{1}{2}gt^2$ * $v_y^2 = u_y^2 + 2a_y y = (u sin\theta)^2 - 2gy$

3. Key Parameters for Ground-to-Ground Projection:

* Time of Flight (T): Total time in air until $y=0$. $T = \frac{2u sin\theta}{g}$ * Maximum Height (H): Occurs when $v_y = 0$. $H = \frac{u^2 sin^2\theta}{2g}$ * Horizontal Range (R): Total horizontal distance. $R = \frac{u^2 sin 2\theta}{g}$

4. Special Cases & Properties:

* Maximum Range: Achieved when $heta = 45^circ$. $R_{max} = \frac{u^2}{g}$. * Complementary Angles: Angles $heta$ and $(90^circ - \theta)$ give the same range for a given $u$. * Velocity at Maximum Height: Only horizontal component $v_x = u cos\theta$ is present.

$v_y = 0$. * Trajectory Equation: $y = x \tan\theta - \frac{gx^2}{2u^2 cos^2\theta}$ (a parabola). * Horizontal Projection from Height 'h': $u_y = 0$. Time to fall $t = sqrt{\frac{2h}{g}}$. Horizontal range $R = u_x t = u_x sqrt{\frac{2h}{g}}$.

5. Relative Motion of Projectiles:

* If two projectiles are launched under gravity, their relative acceleration is zero ($a_{rel} = a_1 - a_2 = g - g = 0$). * Therefore, their relative velocity is constant ($v_{rel} = \text{constant}$), and one projectile appears to move in a straight line relative to the other.

6. Common Traps:

* Confusing $sin\theta$ with $sin 2\theta$ or $sin^2\theta$. * Incorrectly assuming $v=0$ at max height (only $v_y=0$). * Algebraic errors in formula manipulation. * Not resolving initial velocity into components correctly.

Practice problem-solving with these formulas and concepts. Remember to use $g=10,\text{m/s}^2$ unless specified otherwise.

Vyyuha Quick Recall

To remember the key formulas for Projectile Motion:

Time: Two Under Sin Gravity ($T = \frac{2u sin\theta}{g}$) Height: Half Under Sin Square Gravity ($H = \frac{u^2 sin^2\theta}{2g}$) Range: Really Under Sin Two Gravity ($R = \frac{u^2 sin 2\theta}{g}$)

(Think of 'Under' as division, 'Sin Square' as $sin^2\theta$, 'Sin Two' as $sin 2\theta$)