1. Fundamentals and Assumptions
Projectile motion describes the movement of an object (a projectile) near the Earth's surface under the influence of gravity alone.
Key Assumptions (Ideal Projectile Motion)
- Air resistance is negligible.
- The acceleration due to gravity ($g$) is constant in magnitude and direction throughout the motion.
- The Earth's rotation is ignored.
Initial Velocity Decomposition
If a projectile is launched from the origin $(0, 0)$ with an initial speed $v_0$ at an angle $\theta_0$ above the horizontal, the velocity must be decomposed into its independent components.
2. Kinematics and Independent Motion
The motion is analyzed independently in the horizontal ($x$) and vertical ($y$) directions.
2.1. Horizontal Motion
Since air resistance is ignored, there is no force acting in the horizontal direction. Thus, the acceleration is zero, and the velocity is constant.
Derivation: Horizontal Position and Velocity
Given $a_x = 0$, we use the constant acceleration kinematic equations:
Velocity: $v_x(t) = v_{0x} + a_x t$
$$\implies v_x(t) = v_{0x} = v_0 \cos \theta_0$$Position: $x(t) = x_0 + v_{0x} t + \frac{1}{2} a_x t^2$. Assuming $x_0 = 0$:
$$\implies x(t) = (v_0 \cos \theta_0) t$$Conclusion: The horizontal velocity is constant throughout the flight.
2.2. Vertical Motion
The only acceleration is due to gravity, $g$, acting downwards. We define the positive y-direction as upward, so $a_y = -g$.
Derivation: Vertical Position and Velocity
Given $a_y = -g$, we use the constant acceleration kinematic equations. Assuming $y_0 = 0$:
Velocity: $v_y(t) = v_{0y} + a_y t$
$$\implies v_y(t) = v_0 \sin \theta_0 - g t$$Position: $y(t) = y_0 + v_{0y} t + \frac{1}{2} a_y t^2$.
$$\implies y(t) = (v_0 \sin \theta_0) t - \frac{1}{2} g t^2$$Conclusion: The vertical motion is symmetric and follows parabolic motion.
3. Key Projectile Parameters (Derivations)
3.1. Maximum Height ($H$)
The maximum height is reached when the vertical velocity is momentarily zero ($v_y = 0$).
Derivation for Maximum Height
1. Find the time $t_H$ when $v_y = 0$. Use $v_y = v_{0y} - g t$:
$$0 = v_0 \sin \theta_0 - g t_H \implies t_H = \frac{v_0 \sin \theta_0}{g}$$2. Substitute $t_H$ into the vertical position equation $y(t) = v_{0y} t - \frac{1}{2} g t^2$:
$$H = (v_0 \sin \theta_0) \left(\frac{v_0 \sin \theta_0}{g}\right) - \frac{1}{2} g \left(\frac{v_0 \sin \theta_0}{g}\right)^2$$ $$H = \frac{v_0^2 \sin^2 \theta_0}{g} - \frac{1}{2} \frac{v_0^2 \sin^2 \theta_0}{g}$$ $$\implies H = \frac{v_0^2 \sin^2 \theta_0}{2g}$$3.2. Time of Flight ($T$)
The time of flight is the total time the projectile is in the air. For a launch from level ground ($y=0$), the time $T$ is found when $y(T) = 0$.
Derivation for Time of Flight
Set $y(T) = 0$ in the position equation:
$$0 = (v_0 \sin \theta_0) T - \frac{1}{2} g T^2$$Factor out $T$ (since $T \ne 0$ for the flight):
$$T \left(v_0 \sin \theta_0 - \frac{1}{2} g T\right) = 0$$Solve for $T$ from the term in parentheses:
$$v_0 \sin \theta_0 = \frac{1}{2} g T \implies T = \frac{2 v_0 \sin \theta_0}{g}$$Note: $T = 2 t_H$, confirming the symmetry of the flight path.
3.3. Horizontal Range ($R$)
The range is the maximum horizontal distance traveled when the projectile returns to the launch height ($x(T)$).
Derivation for Horizontal Range
Substitute the Time of Flight $T$ into the horizontal position equation $R = x(T) = (v_0 \cos \theta_0) T$:
$$R = (v_0 \cos \theta_0) \left(\frac{2 v_0 \sin \theta_0}{g}\right)$$ $$R = \frac{v_0^2 (2 \sin \theta_0 \cos \theta_0)}{g}$$Using the trigonometric identity $\sin (2\theta) = 2 \sin \theta \cos \theta$:
$$\implies R = \frac{v_0^2 \sin(2 \theta_0)}{g}$$Conclusion: Maximum range occurs when $\sin(2\theta_0) = 1$, so $2\theta_0 = 90^\circ$, or $\theta_0 = 45^\circ$.
3.4. Equation of Trajectory ($y(x)$)
Eliminate the time variable $t$ from the $x(t)$ and $y(t)$ equations to get $y$ as a function of $x$.
Derivation for Trajectory
1. Solve for $t$ from the horizontal equation $x = (v_0 \cos \theta_0) t$:
$$t = \frac{x}{v_0 \cos \theta_0}$$2. Substitute this $t$ into the vertical equation $y = (v_0 \sin \theta_0) t - \frac{1}{2} g t^2$:
$$y(x) = (v_0 \sin \theta_0) \left(\frac{x}{v_0 \cos \theta_0}\right) - \frac{1}{2} g \left(\frac{x}{v_0 \cos \theta_0}\right)^2$$Simplify the first term using $\frac{\sin \theta_0}{\cos \theta_0} = \tan \theta_0$:
$$\implies y(x) = (\tan \theta_0) x - \left(\frac{g}{2 v_0^2 \cos^2 \theta_0}\right) x^2$$This equation is of the form $y = Ax - Bx^2$, which confirms the parabolic path.
4. Practice Problems: Statements (Total: 10)
Use $g = 9.8 \text{ m/s}^2$ unless otherwise stated. Attempt all problems below before checking the solutions.
Intermediate Level (5 Problems)
- A golf ball is launched from ground level with an initial velocity of $30\text{ m/s}$ at an angle of $37^\circ$ above the horizontal. Calculate the time it takes to reach the peak of its flight.
- A stone is thrown horizontally from the top of a cliff $45\text{ m}$ high. If the stone lands $60\text{ m}$ from the base of the cliff, what was the stone's initial speed?
- An archer shoots an arrow with an initial speed of $40\text{ m/s}$. Find the angle $\theta_0$ (in degrees) required to hit a target $120\text{ m}$ away horizontally (assuming the target is at the same height as the launch). Find both possible angles.
- A tennis ball is hit with a velocity of $15\hat{i} + 20\hat{j}\text{ m/s}$ (where $\hat{i}$ is horizontal and $\hat{j}$ is vertical). Determine the maximum height $H$ reached by the ball.
- A projectile is fired at an angle $\theta_0$. If the maximum height $H$ achieved is equal to the horizontal range $R$, what must the launch angle $\theta_0$ be?
Advanced Level (3 Problems)
- A pilot flying horizontally at a constant speed of $250\text{ m/s}$ at an altitude of $500\text{ m}$ drops a supply package. How far horizontally from the drop point will the package land?
- A soccer player kicks a ball with $v_0 = 20\text{ m/s}$ at an angle of $53^\circ$. A second player, located $40\text{ m}$ away in the direction of the kick, starts running toward the ball at the instant it is kicked. What constant speed must the second player run at to intercept the ball just before it hits the ground?
- A projectile is launched from a height $y_0 = 10\text{ m}$ above the ground with $v_0 = 15\text{ m/s}$ and $\theta_0 = 30^\circ$. Find the magnitude of the final velocity of the projectile just before it hits the ground ($y=0$).
Irodov-like Level (2 Problems)
- A particle is projected from the ground at an angle $\theta_0$. At the highest point of its trajectory, the angle of elevation of the particle as seen from the point of projection is $\phi$. Find the relation between the launch angle $\theta_0$ and the angle of elevation $\phi$.
- A stone is thrown with initial velocity $v_0$ at an angle $\alpha$ to the horizontal from the base of a smooth hill. If the hill's slope is inclined at an angle $\beta$ to the horizontal ($\alpha > \beta$), derive the expression for the range $R_{\text{incline}}$ of the stone along the inclined plane.
5. Practice Problems: Solutions (10 Problems)
Detailed solutions for all 10 problems. Use $g=9.8 \text{ m/s}^2$ unless otherwise stated.
Solution P1 (Intermediate)
The time to reach the peak ($t_H$) is found when $v_y = 0$.
Initial vertical velocity: $v_{0y} = v_0 \sin \theta_0 = (30\text{ m/s}) \sin(37^\circ) \approx 18.05\text{ m/s}$.
Using $v_y = v_{0y} - g t_H$, with $v_y=0$:
$$t_H = \frac{v_{0y}}{g} = \frac{18.05\text{ m/s}}{9.8\text{ m/s}^2} \approx 1.84\text{ s}$$Result: The time to reach the peak is approximately $1.84\text{ seconds}$.
Solution P2 (Intermediate)
The stone is thrown horizontally, so $\theta_0 = 0$ and $v_{0y} = 0$. We need to find the time of flight $T$ first using vertical motion, then use horizontal motion to find $v_0$.
1. Time of Flight (Vertical Motion): $y = y_0 + v_{0y} t - \frac{1}{2} g t^2$. With $y_0=45\text{ m}$, $y=0$, $v_{0y}=0$:
$$0 = 45 - 0 - \frac{1}{2} (9.8) T^2 \implies 4.9 T^2 = 45$$ $$T = \sqrt{\frac{45}{4.9}} \approx 3.03\text{ s}$$2. Initial Speed (Horizontal Motion): $x = v_{0x} T$. With $x=60\text{ m}$ and $v_{0x}=v_0$:
$$v_0 = \frac{x}{T} = \frac{60\text{ m}}{3.03\text{ s}} \approx 19.80\text{ m/s}$$Result: The initial speed was approximately $19.80\text{ m/s}$.
Solution P3 (Intermediate)
We use the range formula $R = \frac{v_0^2 \sin(2 \theta_0)}{g}$ and solve for $\sin(2\theta_0)$.
$$120\text{ m} = \frac{(40\text{ m/s})^2 \sin(2 \theta_0)}{9.8\text{ m/s}^2}$$ $$\sin(2 \theta_0) = \frac{120 \times 9.8}{1600} = 0.735$$The two solutions for the angle $2\theta_0$ (since $\sin(180^\circ - \phi) = \sin \phi$) are:
Case 1: $2\theta_0 = \arcsin(0.735) \approx 47.33^\circ$
$$\theta_{0, 1} \approx 23.67^\circ$$Case 2: $2\theta_0 = 180^\circ - 47.33^\circ = 132.67^\circ$
$$\theta_{0, 2} \approx 66.33^\circ$$Result: The required angles are approximately $23.7^\circ$ and $66.3^\circ$ (which are complementary angles, summing to $90^\circ$).
Solution P4 (Intermediate)
The maximum height $H$ depends only on the initial vertical velocity $v_{0y}$.
Given $\vec{v}_0 = 15\hat{i} + 20\hat{j}\text{ m/s}$, so $v_{0y} = 20\text{ m/s}$.
We use the time-independent equation for vertical motion, $v_y^2 = v_{0y}^2 - 2 g (y - y_0)$. At $y=H$, $v_y=0$ and $y_0=0$:
$$0^2 = v_{0y}^2 - 2 g H \implies H = \frac{v_{0y}^2}{2g}$$ $$H = \frac{(20\text{ m/s})^2}{2 \times 9.8\text{ m/s}^2} = \frac{400}{19.6} \approx 20.41\text{ m}$$Result: The maximum height reached is approximately $20.41\text{ meters}$.
Solution P5 (Intermediate)
Set $H=R$ using the formulas derived in Section 3.
$$H = \frac{v_0^2 \sin^2 \theta_0}{2g}, \quad R = \frac{v_0^2 \sin(2 \theta_0)}{g}$$ $$\frac{v_0^2 \sin^2 \theta_0}{2g} = \frac{v_0^2 \sin(2 \theta_0)}{g}$$Cancel $v_0^2$ and $g$ from both sides (assuming $v_0 \ne 0$):
$$\frac{\sin^2 \theta_0}{2} = \sin(2 \theta_0)$$Use the double-angle identity $\sin(2 \theta_0) = 2 \sin \theta_0 \cos \theta_0$:
$$\frac{\sin^2 \theta_0}{2} = 2 \sin \theta_0 \cos \theta_0$$Divide by $\sin \theta_0$ (assuming $\theta_0 \ne 0$):
$$\frac{\sin \theta_0}{2} = 2 \cos \theta_0 \implies \frac{\sin \theta_0}{\cos \theta_0} = 4$$ $$\tan \theta_0 = 4$$ $$\theta_0 = \arctan(4) \approx 75.96^\circ$$Result: The launch angle must be approximately $76.0^\circ$.
Advanced Level Solutions
Solution P6 (Advanced)
The package has an initial horizontal velocity $v_0 = 250\text{ m/s}$ and $v_{0y}=0$. It falls from $y_0 = 500\text{ m}$.
1. Find Time of Flight $T$ (Vertical Motion): $y = y_0 - \frac{1}{2} g t^2$. Set $y=0$:
$$0 = 500 - \frac{1}{2} (9.8) T^2 \implies 4.9 T^2 = 500$$ $$T = \sqrt{\frac{500}{4.9}} \approx 10.10\text{ s}$$2. Find Horizontal Distance $x$ (Horizontal Motion): $x = v_0 T$ (since $v_{0x}=v_0$):
$$x = (250\text{ m/s})(10.10\text{ s}) \approx 2525\text{ m}$$Result: The package will land approximately $2525\text{ meters}$ away horizontally.
Solution P7 (Advanced)
The second player must cover a distance $d = 40\text{ m}$ in the exact time $T$ the ball is in the air. The required speed is $v_{\text{player}} = \frac{40 - R}{T}$, where $R$ is the range of the ball if it lands at $x=R$. The player starts $40\text{ m}$ away and runs towards the landing point $R$.
1. Find Time of Flight $T$ for the ball:
$$T = \frac{2 v_0 \sin \theta_0}{g} = \frac{2 (20) \sin(53^\circ)}{9.8} \approx 3.26\text{ s}$$2. Find Horizontal Range $R$ for the ball:
$$R = \frac{v_0^2 \sin(2 \theta_0)}{g} = \frac{(20)^2 \sin(106^\circ)}{9.8} \approx \frac{400 \times 0.961}{9.8} \approx 39.22\text{ m}$$3. Calculate Player's Required Speed: The player starts at $x=40\text{ m}$ and must reach $x=R \approx 39.22\text{ m}$. The distance the player runs is $40\text{ m} - 39.22\text{ m} = 0.78\text{ m}$.
$$v_{\text{player}} = \frac{\text{Distance}}{\text{Time}} = \frac{0.78\text{ m}}{3.26\text{ s}} \approx 0.24\text{ m/s}$$Result: The second player must run at a constant speed of approximately $0.24\text{ m/s}$ towards the ball.
Solution P8 (Advanced)
We need the final velocity $\vec{v}_f = v_{fx}\hat{i} + v_{fy}\hat{j}$. The magnitude is $|\vec{v}_f| = \sqrt{v_{fx}^2 + v_{fy}^2}$.
1. $v_{fx}$ is constant: $v_{fx} = v_0 \cos \theta_0 = 15 \cos(30^\circ) \approx 12.99\text{ m/s}$.
2. Find $v_{fy}$ using $v_{fy}^2 = v_{0y}^2 - 2 g \Delta y$.
Initial vertical velocity: $v_{0y} = 15 \sin(30^\circ) = 7.5\text{ m/s}$.
Displacement $\Delta y = y_{\text{final}} - y_{\text{initial}} = 0 - 10\text{ m} = -10\text{ m}$.
$$v_{fy}^2 = (7.5)^2 - 2 (9.8) (-10) = 56.25 + 196 = 252.25$$ $$v_{fy} = -\sqrt{252.25} \approx -15.88\text{ m/s} \text{ (negative since it is moving downward)}$$3. Final Magnitude:
$$|\vec{v}_f| = \sqrt{(12.99)^2 + (-15.88)^2} = \sqrt{168.74 + 252.25} = \sqrt{420.99} \approx 20.52\text{ m/s}$$Result: The magnitude of the final velocity is approximately $20.52\text{ m/s}$.
Irodov-like Level Solutions
Solution P9 (Irodov-like)
The highest point of the trajectory occurs at $x=R/2$ and $y=H$. The angle of elevation $\phi$ from the origin $(0, 0)$ to this point is given by the tangent of the angle, $\tan \phi = \frac{\text{Height}}{\text{Base}}$.
$$\tan \phi = \frac{H}{R/2} = \frac{2H}{R}$$Substitute the formulas for Maximum Height ($H$) and Range ($R$):
$$H = \frac{v_0^2 \sin^2 \theta_0}{2g}, \quad R = \frac{v_0^2 \sin(2 \theta_0)}{g}$$ $$\tan \phi = \frac{2 \left(\frac{v_0^2 \sin^2 \theta_0}{2g}\right)}{\frac{v_0^2 \sin(2 \theta_0)}{g}}$$Simplify by canceling common terms ($v_0^2/g$):
$$\tan \phi = \frac{\sin^2 \theta_0}{\sin(2 \theta_0)}$$Use the identity $\sin(2 \theta_0) = 2 \sin \theta_0 \cos \theta_0$:
$$\tan \phi = \frac{\sin^2 \theta_0}{2 \sin \theta_0 \cos \theta_0} = \frac{\sin \theta_0}{2 \cos \theta_0}$$ $$\implies \tan \phi = \frac{1}{2} \tan \theta_0$$Result: The relation between the launch angle $\theta_0$ and the angle of elevation $\phi$ to the highest point is $\tan \phi = \frac{1}{2} \tan \theta_0$.
Solution P10 (Irodov-like)
We use the trajectory equation $y(x)$ and find the intersection point with the plane's equation, $y_{\text{plane}} = (\tan \beta) x$. Let $x_f$ be the horizontal distance to the landing point.
1. Equate the two $y$ positions:
$$(\tan \beta) x_f = (\tan \alpha) x_f - \left(\frac{g}{2 v_0^2 \cos^2 \alpha}\right) x_f^2$$2. Solve for $x_f$ (by dividing by $x_f$, since $x_f \ne 0$):
$$\tan \beta = \tan \alpha - \frac{g x_f}{2 v_0^2 \cos^2 \alpha}$$ $$x_f = \frac{2 v_0^2 \cos^2 \alpha}{g} (\tan \alpha - \tan \beta)$$3. Simplify the trigonometric difference using $\tan A - \tan B = \frac{\sin(A-B)}{\cos A \cos B}$:
$$x_f = \frac{2 v_0^2 \cos^2 \alpha}{g} \left(\frac{\sin(\alpha - \beta)}{\cos \alpha \cos \beta}\right) = \frac{2 v_0^2 \cos \alpha \sin(\alpha - \beta)}{g \cos \beta}$$4. The range along the incline $R_{\text{incline}}$ is the hypotenuse of the right triangle formed by $x_f$ and $y_f$, so $R_{\text{incline}} = \frac{x_f}{\cos \beta}$.
$$R_{\text{incline}} = \frac{1}{\cos \beta} \left[\frac{2 v_0^2 \cos \alpha \sin(\alpha - \beta)}{g \cos \beta}\right]$$ $$\implies R_{\text{incline}} = \frac{2 v_0^2 \cos \alpha \sin(\alpha - \beta)}{g \cos^2 \beta}$$Result: The range along the inclined plane is $R_{\text{incline}} = \frac{2 v_0^2 \cos \alpha \sin(\alpha - \beta)}{g \cos^2 \beta}$.