Physics - Motion in a Plane: Projectile Motion Exhaustive Study Guide

Ground-to-Ground Projectile Motion

• Initial Velocity Components:

  • Let the initial velocity be $\mathbf{u}$. If the angle of projection with the horizontal is $\theta$, then:

    • Horizontal component ($X$-axis): $u_x = u \cos(\theta)$

    • Vertical component ($Y$-axis): $u_y = u \sin(\theta)$

• Acceleration Components:

  • $a_x = 0$ (No horizontal acceleration assuming no air resistance).

  • $a_y = -g$ (Acceleration due to gravity acting downwards).

• Horizontal Velocity at Any Instant:

  • Since $a_x = 0$, the horizontal component of velocity remains constant throughout the motion: $v_x = u \cos(\theta)$.

  • This component serves only to move the particle forward.

• Vertical Velocity at Any Instant:

  • The vertical velocity changes due to gravity: $v_y = u \sin(\theta) - gt$.

  • At the peak (top point), the vertical component of velocity is zero: $v_{y, \text{top}} = 0$.

• Key Behavioral Observations:

  • Velocity at Top Point: The net velocity is not zero. At the highest point, $v_{\text{net}} = u \cos(\theta) \hat{i} + 0 \hat{j} = u \cos(\theta)$.

  • Acceleration Direction: At all points in the trajectory, acceleration is directed downwards: $a_y = -g$.

  • Perpendicularity: Velocity and acceleration are perpendicular only at the top point of the trajectory.

  • Symmetry of Time: In the absence of air resistance, the time of ascent equal to the time of descent: $t_{\text{ascent}} = t_{\text{descent}} = \frac{u \sin(\theta)}{g}$.

  • Speed Symmetry: At the same horizontal level, the speed of the particle is the same, though the velocity vectors differ because their directions are different.

  • Angle Symmetry: The angle of landing at the same level is the same as the angle of projection (with respect to the horizontal).

Kinematic Equations for Projectile Motion

• Velocity of the Particle at Any Time $t$:

  • $\mathbf{v} = v_x \hat{i} + v_y \hat{j}$

  • $\mathbf{v} = (u_x + a_x t) \hat{i} + (u_y + a_y t) \hat{j}$

  • $\mathbf{v} = (u \cos(\theta)) \hat{i} + (u \sin(\theta) - gt) \hat{j}$

  • Speed at any time $t$ is the magnitude of the velocity vector: $|\mathbf{v}| = \sqrt{v_x^2 + v_y^2}$.

• Position / Displacement at Any Time $t$:

  • $\mathbf{s} = s_x \hat{i} + s_y \hat{j}$

  • $s_x = u_x t + \frac{1}{2}a_x t^2 = u \cos(\theta) t$

  • $s_y = u_y t + \frac{1}{2}a_y t^2 = u \sin(\theta) t - \frac{1}{2}gt^2$

  • To find the position at any time, calculate $s_x$ and $s_y$ separately and express them in vector notation.

Fundamental Formulas of Projectile Motion

• Time of Flight ($T$):

  • $T = \frac{2 u_y}{g} = \frac{2 u \sin(\theta)}{g}$

• Maximum Height ($H$):

  • $H = \frac{u_y^2}{2g} = \frac{u^2 \sin^2(\theta)}{2g}$

• Horizontal Range ($R$):

  • $R = \frac{2 u_x u_y}{g} = \frac{u^2 \sin(2\theta)}{g}$

Conditions for Maximum Range

• Maximum Range Condition:

  • Horizontal Range $R = \frac{u^2 \sin(2\theta)}{g}$.

  • For $R$ to be maximum, $\sin(2\theta)$ must be maximum ($1$).

  • $\sin(2\theta) = 1 \Rightarrow 2\theta = 90^\circ \Rightarrow \theta = 45^\circ$.

  • $R_{\text{max}} = \frac{u^2}{g}$.

• Height at Maximum Range:

  • When $\theta = 45^\circ$, $H = \frac{u^2 \sin^2(45^\circ)}{2g} = \frac{u^2 (1/2)}{2g} = \frac{u^2}{4g}$.

  • Relation: $H = \frac{R_{\text{max}}}{4}$.

• Kinetic Energy at the Top:

  • Initial Kinetic Energy ($KE_i$): $KE_i = \frac{1}{2} m u^2$.

  • Velocity at top: $v_{\text{top}} = u \cos(\theta)$.

  • Kinetic Energy at top ($KE_{\text{top}}$): $KE_{\text{top}} = \frac{1}{2} m (u \cos(\theta))^2 = KE_i \cos^2(\theta)$.

Complementary Angles of Projection

• Definition: Angles $\theta$ and $(90^\circ - \theta)$ are complementary.

• Range Equality: The range is the same for two angles $\theta_1$ and $\theta_2$ if they are complementary ($\theta_1 + \theta_2 = 90^\circ$).

  • Example: Projecting at $15^\circ$ and $75^\circ$ results in the same range.

  • Example: Projecting at $(45^\circ + α)$ and $(45^\circ - α)$ results in the same range.

• Height Relations:

  • $H_\theta = \frac{u^2 \sin^2(\theta)}{2g}$

  • $H_{90-\theta} = \frac{u^2 \cos^2(\theta)}{2g}$

  • Ratio of heights: $\frac{H_\theta}{H_{90-\theta}} = \tan^2(\theta)$.

  • Sum of heights: $H_\theta + H_{90-\theta} = \frac{u^2}{2g}$ (This is equal to the height if the object were projected vertically upward with speed $u$).

  • Product of heights and Range: $H_\theta \times H_{90-\theta} = \frac{u^4 \sin^2(\theta) \cos^2(\theta)}{4g^2} = \frac{(u^2 \sin(2\theta))^2}{16g^2} = \frac{R^2}{16}$.

  • Therefore, $R = 4\sqrt{H_\theta H_{90-\theta}}$.

• Time of Flight Relations:

  • $T_\theta = \frac{2 u \sin(\theta)}{g}$

  • $T_{90-\theta} = \frac{2 u \cos(\theta)}{g}$

  • Ratio of times: $\frac{T_\theta}{T_{90-\theta}} = \tan(\theta)$.

  • Product of times: $T_\theta T_{90-\theta} = \frac{4 u^2 \sin(\theta) \cos(\theta)}{g^2} = \frac{2u^2 \sin(2\theta)}{g^2} = \frac{2R}{g}$.

Comprehensive Numerical Examples

• Scenario 1: Height at different angles

  • A ball projected at $30^\circ$ attains height $H_{30} = 50\,m$. Find height at $60^\circ$.

  • Using formula: $\frac{H_{30}}{H_{60}} = \tan^2(30^\circ) = (\frac{1}{\sqrt{3}})^2 = \frac{1}{3}$.

  • $H_{60} = 3 \times H_{30} = 3 \times 50 = 150\,m$.

  • If projected vertically upward at the same speed: $H_v = H_{30} + H_{60} = 50 + 150 = 200\,m$.

• Scenario 2: Detailed Kinematics Analysis

  • Given: $u = 50\sqrt{2}\,m/s$, $\theta = 45^\circ$, $g = 10\,m/s^2$.

  • Components: $u_x = 50\sqrt{2} \cos(45^\circ) = 50\,m/s$; $u_y = 50\sqrt{2} \sin(45^\circ) = 50\,m/s$.

  • Time to reach top: $t = \frac{u_y}{g} = \frac{50}{10} = 5\,s$.

  • Total time of flight: $T = 2 \times 5 = 10\,s$.

  • Maximum Height: $H = \frac{u_y^2}{2g} = \frac{50^2}{2(10)} = 125\,m$.

  • Horizontal Range: $R = u_x T = 50 \times 10 = 500\,m$.

  • Velocity at $t=2\,s$:

    • $v_x = 50\,m/s$

    • $v_y = 50 - 10(2) = 30\,m/s$

    • $\mathbf{v} = 50 \hat{i} + 30 \hat{j}\,m/s$

    • Speed: $\sqrt{50^2 + 30^2}\,m/s$

    • Angle: $\alpha = \tan^{-1}(\frac{30}{50}) = \tan^{-1}(0.6)$.

  • Velocity at $t=10\,s$ (landing):

    • $v_x = 50\,m/s$

    • $v_y = 50 - 10(10) = -50\,m/s$

    • $\mathbf{v} = 50 \hat{i} - 50 \hat{j}\,m/s$

  • Momentum Change (Mass = $10\,kg$):

    • Initial Momentum (at $t=0$): $\mathbf{p}_i = 10(50\hat{i} + 50\hat{j}) = 500\hat{i} + 500\hat{j}\,kg\,m/s$.

    • At $t=5\,s$ (top): $\mathbf{p}_f = 10(50\hat{i} + 0\hat{j}) = 500\hat{i}$.

    • Change in momentum ($\Delta \mathbf{p}$): $\mathbf{p}_f - \mathbf{p}_i = -500\hat{j}\,kg\,m/s$.

  • Displacement at $t=6\,s$:

    • $s_x = 50 \times 6 = 300\,m$

    • $s_y = 50(6) - \frac{1}{2}(10)(6)^2 = 300 - 180 = 120\,m$

    • $\mathbf{s} = 300\hat{i} + 120\hat{j}$.

Questions & Discussion

• Question 11: The ratio of the speed of a projectile at the point of projection to the speed at the top of its trajectory is $x$. Find the angle of projection.

  • Ratio $x = \frac{u}{u \cos(\theta)} = \frac{1}{\cos(\theta)}$.

  • $\cos(\theta) = \frac{1}{x} \Rightarrow \theta = \cos^{-1}(\frac{1}{x})$ (Correct option: D).

• Question 12: Two particles are projected with the same speed at angles $(45^\circ + θ)$ and $(45^\circ - θ)$. Are their ranges same?

  • Yes, because $(45^ο + θ) + (45^ο - θ) = 90^ο$. They are complementary.

• Question 13: If the angle of projection is doubled keeping speed the same and the particle strikes the same target, find the ratio of time of flight.

  • Same target means Same Range, so angles must be $\theta$ and $(90 - \theta)$.

  • Doubled angle implies $2\theta = 90 - \theta \Rightarrow 3\theta = 90 \Rightarrow \theta = 30^ο$.

  • Angles are $30^ο$ and $60^ο$.

  • Ratio of $T = \frac{\sin(30^ο)}{\sin(60^ο)} = \frac{1/2}{\sqrt{3}/2} = 1:\sqrt{3}$ (Correct option: D).

• Question 14: For trajectories A, B, and C shown in a diagram where heights are equal but ranges differ (R_C > R_B > R_A):

  • Since heights are same, $u_y$ is same for all, hence time of flight ($T = \frac{2u_y}{g}$) is same for all.

  • Range depends on $u_x$. Since $R_C$ is largest, particle C has the largest horizontal velocity component.

• Question 17: Maximum area of ground covered by bullets fired in all directions with initial velocity $u$.

  • Area is a circle with radius $R_{\text{max}}$.

  • $R_{\text{max}} = \frac{u^2}{g}$.

  • $Area = \pi R_{\text{max}}^2 = \pi (\frac{u^2}{g})^2 = \pi \frac{u^4}{g^2}$.

• Question 21: Relationship between Range and heights $h_1, h_2$ for the same range.

  • $R = 4\sqrt{h_1 h_2} \Rightarrow R^2 = 16h_1 h_2$ (Correct option: B).

• Question 23: Comparing areas covered by guns with speeds $1\,km/s$ and $2\,km/s$.

  • Area $\propto u^4$.

  • Ratio of areas = $(\frac{1}{2})^4 = \frac{1}{16}$ (Correct option: B).

• Excluded Topics explicitly mentioned as not covered here:

  • Torque

  • Angular Momentum

  • Dot product