UK A-Level: Projectiles

Understanding projectile motion is essential for A-Level Mechanics because it models countless real-world phenomena, from a footballer kicking a ball to an engineer calculating a rocket's trajectory. This topic beautifully illustrates the power of resolving complex two-dimensional motion into simpler, independent one-dimensional components, a core skill in physics that enhances your problem-solving toolkit.

Resolving Initial Velocity into Components

Every projectile problem begins with a launch, characterized by an initial velocity $u$ at an angle $\theta$ to the horizontal. The first critical step is to resolve this single vector into its horizontal and vertical components. This is done using basic trigonometry: the horizontal component is $u_x=u\cos \theta$ and the vertical component is $u_y=u\sin \theta$. Remember, $\theta$ is always measured from the horizontal axis.

Why is this resolution so important? It allows you to treat the motion as two separate, simultaneous events: constant velocity horizontally and constant acceleration vertically. For instance, if a cannonball is fired at 20 m/s at 30° above the ground, its initial horizontal speed is $20\cos 30°\approx 17.3$ m/s, and its initial vertical speed is $20\sin 30°=10$ m/s. This separation is the foundation for all subsequent analysis.

Independent Horizontal and Vertical Motion Equations

With the components established, you apply the equations of motion (suvat equations) to each direction independently. The key assumption is that horizontal and vertical motions are independent; gravity only affects the vertical component. For the horizontal (x-direction) motion, acceleration is zero, so the velocity remains constant: $v_x=u_x=u\cos \theta$. The horizontal displacement is simply: $$s_x=u_{x}t=(u\cos \theta )t$$

For the vertical (y-direction) motion, acceleration is due to gravity, $a_y=-g$, where $g=9.8$ m/s² (taking upward as positive). The standard suvat equations apply: $$v_y=u\sin \theta -gt$$ $$s_y=(u\sin \theta )t-\frac{1}{2}g{t}^2$$ $$v_y^2=(u\sin \theta {)}^2-2g{s}_y$$

You must always define a positive direction (typically upwards) and consistently apply the sign for $g$. A common application is finding when the projectile reaches a certain height by solving the vertical displacement equation for time $t$.

Calculating Time of Flight, Maximum Height, and Range

These three quantities are the hallmark outputs of any standard projectile problem. Their derivations rely on the vertical motion equations and symmetry.

• Time of Flight: This is the total time the projectile spends in the air. For launch and landing at the same vertical level, the vertical displacement $s_y=0$. Using $s_y=(u\sin \theta )t-\frac{1}{2}g{t}^2=0$, you factor out $t$ to find two solutions: $t=0$ (launch) and the time of flight $T=\frac{2u\sin \theta }{g}$.

• Maximum Height: At the peak of its trajectory, the vertical velocity becomes zero ($v_y=0$). Using $v_y^2=(u\sin \theta {)}^2-2g{s}_y$ and setting $v_y=0$, you solve for the vertical displacement $s_y=H$:

$$H=\frac{(u\sin \theta {)}^2}{2g}$$ The time to reach this height is $t=\frac{u\sin \theta }{g}$, exactly half the total time of flight for a level landing.

• Range: The horizontal distance covered, $R$, is found by combining horizontal motion with the total time of flight: $R=u_x\times T=(u\cos \theta )\times \frac{2u\sin \theta }{g}$. This simplifies to:

$$R=\frac{u^2\sin 2\theta }{g}$$ This formula shows the range is maximized when $\sin 2\theta =1$, or at a launch angle of $\theta =45°$.

Consider a worked example: A ball is kicked with an initial speed of 15 m/s at 40° to the horizontal. Find its time of flight, maximum height, and range.

1. Resolve: $u_x=15\cos 40°$, $u_y=15\sin 40°$.
2. Time of flight: $T=\frac{2\times 15\sin 40°}{9.8}\approx \frac{19.28}{9.8}\approx 1.97$ s.
3. Maximum height: $H=\frac{(15\sin 40°{)}^2}{2\times 9.8}\approx \frac{(9.64{)}^2}{19.6}\approx 4.74$ m.
4. Range: $R=(15\cos 40°)\times 1.97\approx 11.49\times 1.97\approx 22.6$ m, or directly using $R=\frac{15^2\sin 80°}{9.8}\approx 22.6$ m.

Projectiles on Inclined Planes and Varied Angles

Real-world launches often don't start and end on the same horizontal level. You might project an object up or down a slope, or from a cliff. The core principles remain the same, but the geometry becomes more involved.

For varied launch and landing heights, you cannot use the simplified formulas for $T$ and $R$. Instead, you must use the general vertical displacement equation $s_y=(u\sin \theta )t-\frac{1}{2}g{t}^2$, where $s_y$ is the net vertical displacement (e.g., -10 m if landing 10 m below the launch point). You solve this quadratic for $t$ to find the time of flight, then use $s_x=(u\cos \theta )t$ for the horizontal range.

Inclined plane problems introduce a new coordinate system. Instead of horizontal and vertical, you resolve vectors parallel and perpendicular to the plane. For example, if a projectile is launched from a plane inclined at angle $\alpha$, the initial velocity components become:

• Parallel to plane: $u_{\parallel }=u\cos (\theta -\alpha )$
• Perpendicular to plane: $u_{\perp }=u\sin (\theta -\alpha )$

The acceleration due to gravity also has components: $g\sin \alpha$ down the plane and $g\cos \alpha$ into the plane. You then apply suvat equations in these two new directions. A typical question asks for the range up the slope, which involves finding the time when the perpendicular displacement returns to zero (hitting the slope) and then calculating the parallel displacement at that time.

Common Pitfalls

1. Mixing Horizontal and Vertical Quantities: The most frequent error is using a vertical time in a horizontal equation without ensuring it's the correct instant. Remember, time $t$ is the common variable linking the two motions, but values like maximum height time and total flight time are specific. Always clearly define which time value you are using in each equation.

1. Ignoring the Launch and Landing Conditions: Forgetting that $s_y=0$ only applies for level ground can derail a problem. If a ball is thrown from a 5m high cliff, the landing condition is $s_y=-5$ m (if up is positive). Always write down the known displacement values for your chosen coordinate system before applying equations.

1. Sign Errors with Gravity: Consistency is key. If you define upwards as positive, then acceleration $a_y=-9.8$ m/s². This negative sign must be included in all vertical suvat equations. A missed sign will give incorrect results for time, height, or velocity.

1. Misapplying the Range Formula: The formula $R=\frac{u^2\sin 2\theta }{g}$ is only valid for launch and landing at the same horizontal level. Using it for problems with a height difference or an inclined plane will yield a wrong answer. Always revert to the fundamental component equations when conditions are not standard.

Summary

• Projectile motion is analyzed by resolving the initial velocity into independent horizontal ($u\cos \theta$) and vertical ($u\sin \theta$) components.
• Horizontal motion has constant velocity, while vertical motion has constant acceleration due to gravity ($a_y=-g$), described by the suvat equations.
• Key results for level ground include time of flight $T=\frac{2u\sin \theta }{g}$, maximum height $H=\frac{(u\sin \theta {)}^2}{2g}$, and range $R=\frac{u^2\sin 2\theta }{g}$ (maximized at 45°).
• For non-level ground or inclined planes, success depends on carefully defining the coordinate system, applying the correct launch/landing conditions, and solving the general motion equations.
• Always treat time as the linking variable and maintain strict sign conventions for displacement, velocity, and acceleration throughout your calculations.