Projectile Motion
Projectile motion describes how objects move through the air after being launched, like when a ball is thrown or an arrow fired. After launching, the only force acting on the projectile is gravity (ignoring effects of air resistance for simplicity). This produces a parabolic trajectory unless launched vertically.
Use this page to revise the following concepts within projectile motion:
One-dimensional motion
One-dimensional projectile motion occurs when an object is launched vertically. Since it moves in a straight line under constant gravitational acceleration, its motion can be analysed using kinematic equations (SUVAT).
As gravity is the only force acting on the object after launch, the vertical motion exhibits symmetry when projected upwards. The time taken to reach the highest point is equal to the time taken to descend back to the launch height. This symmetry also applies to velocity : at any given height during ascent, the object's speed (in magnitude) matches its speed at the same height during descent, but in the opposite direction.
This is illustrated in the diagram below where a ball is being thrown upwards, reaching a peak and then returning back down. \(t_2\) occurs at the same time after \(t_{\text{peak}}\) as \(t_1\) occurs before it, showing a symmetrical point where the velocity is the same but in the opposite direction.
Worked Example
A ball is thrown vertically upwards with an initial velocity of \(15 \mathrm{~ms}^{-1}\) from the ground. Ignore air resistance and use \(g = 9.8 \mathrm{~ms}^{-2}\).
Calculate:
- The time taken to reach the highest point
- The maximum height reached
- The total time before the ball reaches the ground.
Define the positive direction: upwards.
Consider the case from when the ball is initially thrown to when it reaches its maximum height. Identify the known variables:
\[ \begin{aligned} s &= ? \\ u &= 15 \mathrm{~ms}^{-1} \\ v &= 0 \mathrm{~ms}^{-1} \text{ at the peak} \\ a &= -9.8 \mathrm{~ms}^{-2} \\ t &= ? \end{aligned} \]
Determine the appropriate equation to use to find the time taken:
\[ \begin{aligned} v &= u + at \\ t &= \frac{v - u}{a} \end{aligned} \]
Substitute variables:
\[ \begin{aligned} t &= \frac{0 - 15}{-9.8} \\ &\approx 1.53 \mathrm{~s} \end{aligned} \]
Determine the appropriate equation to find the maximum height reached:
\[ \begin{aligned} v^2 &= u^2 + 2as \\ s &= \frac{v^2 - u^2}{2a} \end{aligned} \]
Substitute variables:
\[ \begin{aligned} s &= \frac{0 - 15^2}{2(-9.8)} \\ &\approx 11.48 \mathrm{~m} \end{aligned} \]
As the motion is symmetrical, the time for the ball to reach the peak and the time for the ball to come back down from the peak are equal. Therefore, the total time can be calculated as:
\[ \begin{aligned} t_{\text{total}} &= 2t \\ &= 3.06 \mathrm{~s} \end{aligned} \]
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Two-dimensional motion
When an object is launched at any angle other than vertically, it follows a parabolic trajectory. To analyse this motion, we break it into two independent components: vertical and horizontal motion.
- Vertical motion: The only force acting on the projectile is gravity, meaning its motion follows the same principles as one-dimensional projectile motion. The projectile accelerates downwards at \(9.8\text{ ms}^{-2}\) reaching its peak height when its vertical velocity becomes zero before descending symmetrically.
- Horizontal motion: Since air resistance is ignored, no horizontal forces act on the projectile, meaning it moves at a constant velocity throughout its flight.
The time of flight is determined by the vertical motion, as the projectile remains in the air until it returns to its original launch height. Using this, we can calculate the range, or horizontal distance travelled, based on the time in the air and the constant horizontal velocity.
This is illustrated in the diagram below, where:
- The blue arrow represents the velocityvector (which points tangential to the projectile path)
- The green arrow shows the horizontal component of velocity (which stays constant throughout)
- The red arrow indicates the vertical component of velocity (which changes due to gravity)
Worked example
An archer releases an arrow from a height of \(1.6 \mathrm{~m}\) with an initial velocity of \(45 \mathrm{~ms}^{-1}\) at an angle of \(30^\circ\) above the horizontal. The target is placed on a stand at the same height as the arrow's release point \((1.6 \mathrm{~m})\).
Assuming the archer hits the target, there is no air resistance, and \(g = 9.8 \mathrm{~ms}^{-2}\), find:
- How much time passes before the arrow reaches the target.
- The maximum height reached by the arrow above the release point.
- The horizontal distance to the target.
Define the positive directions: upwards and to the right.
Identify the known variables:
\(u = 45 \mathrm{~ms}^{-1}\)
\(\theta = 30^\circ\)
\(a = -9.8 \mathrm{~ms}^{-2}\)
Find the horizontal component \(u_h\) and vertical component \(u_v\) of the initial velocity:
\[ \begin{aligned} u_h &= u\cos(\theta) = 45\cos(30^\circ) \approx 38.97 \mathrm{~ms}^{-1} \\ u_v &= u\sin(\theta) = 45\sin(30^\circ) = 22.5 \mathrm{~ms}^{-1} \end{aligned} \]
As the initial and final heights are the same, the vertical displacement is \(s_v = 0 \mathrm{~m}\).
To find the time taken, use the following vertical motion equation:
\[ s_v = u_vt + \frac{1}{2}at^2 \]
Substitute variables into the equation:
\[ \begin{aligned} 0 &= 22.5t + \frac{1}{2}(-9.8)t^2 \\ 0 &= 22.5t - 4.9t^2 \\ t(22.5 - 4.9t) &= 0 \\ t &= \frac{22.5}{4.9} \\ &\approx 4.59 \mathrm{~s} \end{aligned} \]
To find the maximum height, consider the motion from release to the peak.
At the peak, \(v_v = 0 \mathrm{~ms}^{-1}\).
Identify the correct motion equation:
\[ v_v^2 = u_v^2 + 2as_v \]
Substitute values into the equation:
\[ \begin{aligned} 0 &= (22.5)^2 + 2(-9.8)s_v \\ 19.6s_v &= 506.25 \\ s_v &\approx 25.83 \mathrm{~m} \end{aligned} \]
To find the horizontal distance, use the horizontal motion equation:
\[ s_h = u_ht \]
Substitute values into the equation:
\[ \begin{aligned} s_h &= 38.97(4.59) \\ &\approx 178.88 \mathrm{~m} \end{aligned} \]
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Different cases of projectile motion
Projectile motion scenarios can vary in presentation, but the analysis and problem-solving approach remains consistent:
- Determine the horizontal and vertical components of the initial velocity
- Use the vertical component to determine the time of flight
- Find the peak of the motion (where the vertical velocity is zero)
- Determine the horizontal range using the time of flight (as horizontal velocity is constant)
The scenarios may differ as:
- The projectile lands at a different height than it started (higher or lower).
- The projectile starts from an elevated position (e.g. launched from a hill or platform).
- The projectile is aimed at a target at a different height (e.g. throwing a ball into a hoop).
In these scenarios, analysing the vertical motion in two cases may be helpful (without the use of the quadratic formula):
- Ascent (start to peak): Vertical velocity decreases to zero at the peak.
- Descent (peak to final height): Vertical velocity increases as the projectile falls.
To solve these problems, first determine the height at the peak, then calculate the total displacement for each stage before finding the total time of flight.
Worked example
A golfer stands on a \(4 \mathrm{~m}\) high platform and strikes a golf ball with an initial speed of \(40 \mathrm{~ms}^{-1}\) at an angle of \(15^\circ\) above the horizontal. The ball follows projectile motion before landing on the ground below.
Assuming there is no air resistance, and \(g = 9.8 \mathrm{~ms}^{-2}\), find:
- the maximum height reached by the ball above ground level.
- the total time the ball is in the air.
- the horizontal distance the ball has travelled before hitting the ground.
Define the positive directions: upwards and to the right.
Identify the known variables:
\(u = 40 \mathrm{~ms}^{-1}\)
\(\theta = 15^\circ\)
\(a = -9.8 \mathrm{~ms}^{-2}\)
Find the horizontal component \(u_h\) and vertical component \(u_v\) of the initial velocity:
\[ \begin{aligned} u_h &= u\cos(\theta) = 40\cos(15^\circ) \approx 38.64 \mathrm{~ms}^{-1} \\ u_v &= u\sin(\theta) = 40\sin(15^\circ) \approx 10.35 \mathrm{~ms}^{-1} \end{aligned} \]
As the projectile starts at an initial height, the motion can be analysed in two cases: ascent and descent.
From the platform to the peak, analyse the vertical motion:
At the peak, \(v_v = 0 \mathrm{~ms}^{-1}\).
Identify the correct motion equation:
\[ v_v = u_v + at \]
Substitute values:
\[ \begin{aligned} 0 &= 10.35 + (-9.8)t \\ t &= \frac{10.35}{9.8} \\ &\approx 1.06 \mathrm{~s} \end{aligned} \]
To determine the time taken from the peak to the ground, first find the vertical distance from the platform to the peak:
Identify the correct motion equation:
\[ v_v^2 = u_v^2 + 2as_v \]
Substitute values:
\[ \begin{aligned} 0^2 &= (10.35)^2 + 2(-9.8)s_v \\ s_v &= \frac{(10.35)^2}{2(9.8)} \\ &\approx 5.47 \mathrm{~m} \end{aligned} \]
The total height from ground level to the peak is:
\[ h_{\text{max}} = 5.47 + 4 = 9.47 \mathrm{~m} \]
From the peak to the ground, analyse the vertical motion:
\(s_v = -9.47 \mathrm{~m}\)
\(u_v = 0 \mathrm{~ms}^{-1}\)
Identify the correct motion equation:
\[ s_v = u_vt + \frac{1}{2}at^2 \]
Substitute values:
\[ \begin{aligned} -9.47 &= 0t + \frac{1}{2}(-9.8)t^2 \\ t &= \sqrt{\frac{2(9.47)}{9.8}} \\ &\approx 1.39 \mathrm{~s} \end{aligned} \]
The time from peak to ground is \(1.39 \mathrm{~s}\).
Total flight time is:
\[ t_{\text{total}} = 1.06 + 1.39 = 2.45 \mathrm{~s} \]
To find the horizontal distance, use the horizontal motion equation:
\[ s_h = u_ht \]
Substitute values into the equation:
\[ \begin{aligned} s_h &= (38.64)(2.45) \\ &\approx 94.67 \mathrm{~m} \end{aligned} \]