Projectile Motion (HSC SSCE Physics): Revision Notes

The Trajectory of a Projectile

Introduction to projectile trajectories

When you throw, kick, or launch an object into the air, it follows a curved path called a trajectory. This trajectory is parabolic in shape when air resistance is negligible. A projectile can be launched at any angle $\theta$ and from any initial height above the ground.

The launch height is the initial vertical position from which the projectile is released. You can define your coordinate system in different ways - either setting the launch height as zero and measuring downward, or setting the ground as zero and measuring the launch height as positive.

The diagram above shows two important scenarios:

• Figure a: A projectile launched from ground level ($y_0 = 0$) that lands back at ground level
• Figure b: A projectile launched from above ground level with a positive launch height

Both trajectories show two key measurements:

• Maximum height: The highest point the projectile reaches
• Horizontal range: The total horizontal distance travelled

Maximum height

What is maximum height?

The maximum height ($h$) is the highest vertical position reached by a projectile during its flight. This occurs at the exact moment when the vertical velocity becomes zero - the projectile stops moving upward before it begins to fall back down.

Calculating maximum height

The vertical position of a projectile at any time is given by the kinematic equation:

$y = u_y t + \frac{1}{2}gt^2$

At the maximum height, we know that $v_y = 0$ (vertical velocity is zero) and $y = h$ (vertical position equals maximum height).

Using the equation $v_y = u_y + gt$, we can write:

$0 = u_y + gt$

Rearranging to find the time at which maximum height occurs:

$t = -\frac{u_y}{g}$

Now we can substitute this time back into the position equation:

$h = u_y t + \frac{1}{2}gt^2 = u_y\left(-\frac{u_y}{g}\right) + \frac{1}{2}g\left(-\frac{u_y}{g}\right)^2 = -\frac{u_y^2}{2g}$

If the launch height is not zero, the maximum height is the launch height plus the additional height gained:

$h = y_0 - \frac{u_y^2}{2g}$

Alternative derivation

You can also derive the maximum height formula using the equation $v^2 = u^2 + 2as$. In the vertical direction:

$v_y^2 = u_y^2 + 2gy$

At maximum height, $v_y = 0$ and $y = h$, giving:

$0 = u_y^2 + 2gh$

$h = -\frac{u_y^2}{2g}$

This is exactly the same result, demonstrating the consistency of kinematic equations.

Worked example: Maximum height

Let's work through a complete example to demonstrate the calculation process.

Time of flight

Understanding time of flight

The time of flight is the total time interval between when a projectile is launched and when it hits the ground. Understanding this concept requires us to look carefully at the shape of the trajectory.

For a symmetric trajectory (where the launch height equals the landing height), the path is perfectly parabolic. This symmetry means:

• Time to reach maximum height = Time to fall from maximum height back to ground
• The ascent phase mirrors the descent phase

We already found that the time to reach maximum height is:

$t = -\frac{u_y}{g}$

Therefore, for a symmetric trajectory, the total time of flight is:

$t_{\text{flight}} = -\frac{2u_y}{g}$

Non-symmetric trajectories

When the launch and landing heights are different, it's simplest to consider the flight in two separate phases:

1. Ascent: From launch point to maximum height
2. Descent: From maximum height to landing point

Worked example: Time of flight

Continuing with Jen's textbook thrown from a window, let's find how long it takes to reach the ground.

Final velocity

For symmetric trajectories

When a projectile lands at the same height from which it was launched, something interesting happens. The final velocity has the same magnitude as the initial velocity, but its direction is different.

Why does this occur?

• The horizontal velocity remains constant throughout the flight (no horizontal forces)
• The vertical velocity decreases by the same amount during ascent as it increases during descent
• The path is symmetric, so the velocity is also symmetric

For non-symmetric trajectories

When launch and landing heights differ, it's easiest to calculate the final velocity by finding its components separately.

Horizontal component: This remains constant throughout the flight:

$v_x = u_x = u\cos\theta$

Vertical component: We need to consider only the descent phase. If we know the time for this phase ($t_2$), we can calculate:

$v_y = u_y + gt$

For the descent phase starting from maximum height, $u_y = 0$, so:

$v_y = gt_2$

Once we have both components, we can find the magnitude and direction of the final velocity.

Worked example: Final velocity

Let's find the final velocity of Jen's textbook using the component method.

Horizontal range

Understanding horizontal range

The horizontal range (often simply called "the range") is the total horizontal distance a projectile travels before it lands. You can find it directly from the time of flight, because the horizontal velocity remains constant throughout the flight.

From the kinematic equations:

$x_{\text{max}} = u_x t_{\text{flight}}$

Since $u_x = u\cos\theta$, we can write:

$x_{\text{max}} = u\cos\theta \times t_{\text{flight}}$

Worked example: Horizontal range

How far from the window does Jen's textbook land?

Range formula for symmetric trajectories

For projectiles that land at the same height from which they were launched, we can derive a useful formula. Starting with:

$x_{\text{max}} = u_x t_{\text{flight}}$

Substituting $u_x = u\cos\theta$ and $t_{\text{flight}} = -\frac{2u_y}{g} = -\frac{2u\sin\theta}{g}$:

$x_{\text{max}} = u\cos\theta \times \left(-\frac{2u\sin\theta}{g}\right) = \frac{2u^2\sin\theta\cos\theta}{g}$

The effect of launch angle

An important result from projectile motion theory is that the maximum possible horizontal range occurs when the launch angle is 45° (assuming no air resistance).

The diagram above compares trajectories for the same initial speed but different launch angles: $15°$, $30°$, $45°$, $60°$, and $75°$. Notice that:

• The $45°$ trajectory achieves the greatest horizontal distance
• Angles below $45°$ have flatter trajectories with less maximum height
• Angles above $45°$ have higher maximum heights but shorter ranges
• Complementary angles (like $30°$ and $60°$) achieve the same range

Investigation: Measuring horizontal range

This practical investigation allows you to measure the launch velocity and time of flight of a projectile experimentally and compare your results with theoretical predictions.

Aim

To find the launch velocity and time of flight of a projectile.

Materials

• Curved ramp (such as a toy car track)
• Ball bearing
• Tape measure
• Sand tray

Risk assessment

Risk: The ball bearing may hit someone Management: Keep the area at the end of the ramp clear

Method

1. Set up the curved track on the edge of a desk as shown in the diagram. Ensure the end of the track is horizontal.
2. Measure the vertical distance from the ground to the desk edge ($h_{\text{desk}}$) and from the ground to the top of the track ($h_{\text{track}}$).
3. Release the ball bearing from the top of the track. Observe where it lands and position the sand tray accordingly.
4. Release the ball bearing again from the top of the track.
5. Measure the horizontal distance ($x$) from the edge of the desk to the point where the ball bearing landed in the sand.
6. Smooth the sand and repeat steps 4 and 5 at least five times to get multiple measurements.

Results

Record all measurements:

• Heights: $h_{\text{desk}}$ and $h_{\text{track}}$
• Multiple measurements of horizontal range $x$

Analysis

1. Calculate time of flight: Use $h_{\text{desk}}$ and the equation $h = \frac{1}{2}gt^2$ to find how long the ball is in the air. Assume the ball leaves the track horizontally ($u_y = 0$).

2. Calculate average range: Find the mean of your horizontal range measurements and determine the uncertainty.

3. Calculate initial velocity: Use $x_{\text{max}} = u_x t_{\text{flight}}$ with your average range and time of flight. Since the ball leaves horizontally, $u_x = u$. Calculate the uncertainty in your result.

4. Compare with energy method: Use conservation of energy to predict what the velocity should be based on $h_{\text{track}}$. The potential energy at the top ($mgh_{\text{track}}$) converts to kinetic energy at the bottom ($\frac{1}{2}mu^2$), assuming no energy losses.

Remember!