Projectile and Uniform Circular Motion

Lessons

Senior 4 · Physics

Projectile and Uniform Circular Motion

5 lessons

Lesson 2 of 5

Oblique Projection: Time of Flight, Maximum Height and Range

25 min

The best angle

Point a garden hose straight out sideways and the water lands close by. Tilt it up a little and the jet reaches further; tilt it more and it reaches further still; but tilt it too steeply and the water climbs high and falls back near your feet. Somewhere in between is an angle that sends the water furthest. Footballers taking a long kick, javelin throwers and the engineers who design irrigation sprinklers all face the same question. In this lesson you derive the time of flight, maximum height and range of a projectile launched at any angle, find the best angle, and see what changes when the launch point is above the ground or the air pushes back.

Splitting an oblique launch

A body launched with speed $u$ at an angle $\theta$ above the horizontal is an oblique projectile. Resolve the launch velocity (Figure 2.1):

u_x = u\cos\theta, \qquad u_y = u\sin\theta

With up positive and the launch point as origin, the equations of Lesson 1 become

x = (u\cos\theta)\,t, \qquad y = (u\sin\theta)\,t - \tfrac{1}{2} g t^{2}, \qquad v_x = u\cos\theta, \qquad v_y = u\sin\theta - g t

Figure 2.1: An oblique projectile: the launch velocity u at angle θ split into components; the maximum height H is reached halfway along the range R, where the velocity is horizontal.

Time of flight, maximum height and range

Time to the top. At the highest point the vertical velocity is zero: $0 = u\sin\theta - g t$ , so $t_{\text{top}} = \frac{u\sin\theta}{g}$ .

Maximum height. Using $v_y^{2} = u_y^{2} - 2gy$ with $v_{y} = 0$ at $y = H$ :

H = \frac{u^{2}\sin^{2}\theta}{2g}

Time of flight. Back on level ground $y = 0$ : $t\,(u\sin\theta - \tfrac{1}{2} g t) = 0$ , so apart from $t = 0$ ,

T = \frac{2u\sin\theta}{g} = 2\,t_{\text{top}}

Range. The horizontal distance covered in time $T$ , using $2\sin\theta\cos\theta = \sin 2\theta$ :

R = u\cos\theta \times \frac{2u\sin\theta}{g} = \frac{u^{2}\sin 2\theta}{g}

Equation of the trajectory. Put $t = \frac{x}{u\cos\theta}$ into the equation for $y$ :

y = x\tan\theta - \frac{g\,x^{2}}{2u^{2}\cos^{2}\theta}

This has the form $y = ax - bx^{2}$ : the trajectory is a parabola, symmetric about the vertical line through its highest point. These results hold only when the projectile lands at the height it was launched from and air resistance is negligible.

Definition

For a projectile launched with speed $u$ at angle $\theta$ above the horizontal and landing at the same level (no air resistance):

time of flight $T = \frac{2u\sin\theta}{g}$ ;
maximum height $H = \frac{u^{2}\sin^{2}\theta}{2g}$ ;
horizontal range $R = \frac{u^{2}\sin 2\theta}{g}$ , greatest, $R_{\max} = \frac{u^{2}}{g}$ , when $\theta = 45^\circ$ .

Example: for $u = 10\,\mathrm{m\,s^{-1}}$ at $45^\circ$ , $R_{\max} = \frac{10^{2}}{9.8} = 10.2\,\mathrm{m}$ .

Worked example 1: a long kick

A football is kicked from the ground at $20\,\mathrm{m\,s^{-1}}$ at $30^\circ$ above the horizontal. Find its time of flight, maximum height and range.

$u_x = 20\cos 30^\circ = 17.3\,\mathrm{m\,s^{-1}}$ , $u_y = 20\sin 30^\circ = 10.0\,\mathrm{m\,s^{-1}}$ .

T = \frac{2 \times 10.0}{9.8} = 2.04\,\mathrm{s}, \qquad H = \frac{10.0^{2}}{2 \times 9.8} = 5.10\,\mathrm{m}, \qquad R = 17.3 \times 2.04 = 35.3\,\mathrm{m}

Check with the range formula: $R = \frac{20^{2}\sin 60^\circ}{9.8} = 35.3\,\mathrm{m}$ .

The best angle, and pairs of angles

Since $\sin 2\theta$ is at most 1, when $2\theta = 90^\circ$ , the range is greatest at $\theta = 45^\circ$ . Also $\sin 2\theta = \sin(180^\circ - 2\theta) = \sin 2(90^\circ - \theta)$ , so complementary angles, such as $30^\circ$ and $60^\circ$ , give the same range: the steeper launch goes higher and stays up longer, the flatter one arrives sooner (Figure 2.2).

Figure 2.2: Trajectories at five angles for the same launch speed, 20 m/s, drawn to scale.

Figure 2.3 shows how the quantities change with time for the kick of Worked example 1. The horizontal velocity is constant; the vertical velocity falls steadily with gradient $- g$ and passes through zero at the top; the height rises and falls symmetrically.

Figure 2.3: Graphs of the motion of a ball kicked at 20 m/s at 30°. The gradient of the vertical-velocity graph is −g = −9.8 m s⁻²; the gradient of the height graph is the vertical velocity.

Worked example 2: hitting a target

A water jet leaves a nozzle at $20\,\mathrm{m\,s^{-1}}$ . At what angles must it be aimed to hit a point on the ground 30 m away, at the level of the nozzle?

\sin 2\theta = \frac{gR}{u^{2}} = \frac{9.8 \times 30}{20^{2}} = 0.735

$2\theta = 47.3^\circ$ or $180^\circ - 47.3^\circ = 132.7^\circ$ , so $\theta = 23.7^\circ$ or $66.3^\circ$ . Both angles work. If $\frac{gR}{u^{2}}$ were greater than 1, no angle would reach: the target would be beyond $R_{\max} = \frac{u^{2}}{g} = 40.8\,\mathrm{m}$ .

Worked example 3: launched from above the ground

A shot-putter releases the shot 2.0 m above the ground at $12\,\mathrm{m\,s^{-1}}$ at $40^\circ$ above the horizontal. Find how far away it lands.

$u_x = 12\cos 40^\circ = 9.19\,\mathrm{m\,s^{-1}}$ , $u_y = 12\sin 40^\circ = 7.71\,\mathrm{m\,s^{-1}}$ . It lands where $y = -2.0\,\mathrm{m}$ :

-2.0 = 7.71t - 4.9t^{2} \;\Rightarrow\; 4.9t^{2} - 7.71t - 2.0 = 0 \;\Rightarrow\; t = \frac{7.71 + \sqrt{7.71^{2} + 4 \times 4.9 \times 2.0}}{2 \times 4.9} = 1.80\,\mathrm{s}

(The other root is negative and is rejected.) $R = 9.19 \times 1.80 = 16.5\,\mathrm{m}$ . Launched from ground level at the same speed, even the best angle, $45^\circ$ , gives only $\frac{12^{2}}{9.8} = 14.7\,\mathrm{m}$ . When the release point is above the landing point, the best angle is a little less than $45^\circ$ .

Applications, and the effect of air

The same equations are used to aim water from fire hoses and irrigation sprinklers, to plan the paths of balls in football, basketball and volleyball, to set the angle of a javelin or shot, and to work out where debris from a volcano or a blast can land. In every real case the air pushes back on the moving body. Air resistance opposes the velocity and grows with speed, so it shortens the range, lowers the maximum height, and makes the descent steeper and shorter than the climb: the path is no longer a symmetric parabola. For a light, fast object such as a shuttlecock the effect is huge; for a heavy shot it is small, which is why the simple equations work well for it.

Warning

Common mistakes. (1) Using $R = \frac{u^{2}\sin 2\theta}{g}$ when the landing point is higher or lower than the launch point: use the component equations and solve for $t$ instead. (2) Writing $\sin^{2}\theta$ as $\sin 2\theta$ : they are different. (3) Thinking the velocity is zero at the top: only the vertical component is zero; the body still moves horizontally at $u\cos\theta$ . (4) Leaving the calculator in radians when the angle is in degrees.

Tip

Memory hook: "Up and down share the time, sideways shares the speed." The vertical motion gives the time; multiply by the steady horizontal speed for the range. And remember the pairs: $45^\circ$ is best, and $\theta$ and $90^\circ - \theta$ land in the same place.

Projectile Motion: Maximum Range Angle (The Organic Chemistry Tutor). Watch how the range formula is differentiated with respect to the launch angle to show that 45° gives the greatest range on level ground.

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Practice

A ball is kicked from the ground at $25\,\mathrm{m\,s^{-1}}$ at $37^\circ$ above the horizontal. Find the components of its launch velocity, its time of flight, maximum height and range.
Starting from the component equations, derive $R = \frac{u^{2}\sin 2\theta}{g}$ and show that the range is greatest at $45^\circ$ .
Water leaves a hose at $10\,\mathrm{m\,s^{-1}}$ . What is the furthest horizontal distance it can reach at the level of the nozzle?
A ball is launched at $10\,\mathrm{m\,s^{-1}}$ from level ground and lands 8.0 m away. Find the two possible launch angles.
A projectile is launched at $30\,\mathrm{m\,s^{-1}}$ at $60^\circ$ . Find its speed and direction of motion 2.0 s after launch.
A javelin is released 1.8 m above the ground at $28\,\mathrm{m\,s^{-1}}$ at $35^\circ$ . Ignoring air resistance, how far away does it land?
Sketch, on the same time axis, graphs of $v_{x}$ , $v_{y}$ and $y$ for an oblique projectile, and explain how the gradient of each graph is related to another quantity.
A coach tells players: "Always kick at $45^\circ$ for the longest distance." Evaluate this advice for a real football.

Show answersHide answers

$u_x = 25\cos 37^\circ = 20.0\,\mathrm{m\,s^{-1}}$ , $u_y = 25\sin 37^\circ = 15.0\,\mathrm{m\,s^{-1}}$ ; $T = \frac{2 \times 15.0}{9.8} = 3.07\,\mathrm{s}$ ; $H = \frac{15.0^{2}}{19.6} = 11.5\,\mathrm{m}$ ; $R = 20.0 \times 3.07 = 61.3\,\mathrm{m}$ .
$T = \frac{2u\sin\theta}{g}$ , $R = u\cos\theta \cdot T = \frac{2u^{2}\sin\theta\cos\theta}{g} = \frac{u^{2}\sin 2\theta}{g}$ ; $\sin 2\theta \le 1$ with equality at $2\theta = 90^\circ$ .
$R_{\max} = \frac{10^{2}}{9.8} = 10.2\,\mathrm{m}$ , at $45^\circ$ .
$\sin 2\theta = \frac{9.8 \times 8.0}{10^{2}} = 0.784$ : $2\theta = 51.6^\circ$ or $128.4^\circ$ , so $\theta = 25.8^\circ$ or $64.2^\circ$ .
$v_x = 30\cos 60^\circ = 15.0\,\mathrm{m\,s^{-1}}$ ; $v_y = 30\sin 60^\circ - 9.8 \times 2.0 = 25.98 - 19.6 = 6.38\,\mathrm{m\,s^{-1}}$ ; $v = \sqrt{15.0^{2} + 6.38^{2}} = 16.3\,\mathrm{m\,s^{-1}}$ at $\tan^{-1}\frac{6.38}{15.0} = 23^\circ$ above the horizontal (still rising).
$u_x = 22.9\,\mathrm{m\,s^{-1}}$ , $u_y = 16.1\,\mathrm{m\,s^{-1}}$ ; $-1.8 = 16.06t - 4.9t^{2} \Rightarrow t = 3.39\,\mathrm{s}$ ; $R = 22.9 \times 3.39 = 77.7\,\mathrm{m}$ .
$v_{x}$ : horizontal line (gradient 0, as $a_{x} = 0$ ). $v_{y}$ : straight line falling from $u\sin\theta$ to $-u\sin\theta$ , gradient $- g$ . $y$ : symmetric hump, zero at 0 and $T$ ; its gradient at any time equals $v_{y}$ , zero at the top.
On level ground with no air, $45^\circ$ is best. A real football is slowed by air resistance, which cuts high, long flights most, so the best angle is lower than $45^\circ$ ; spin and wind change it further, and a kick that must stay low (to beat defenders, or keep the ball in play) uses a smaller angle anyway. The advice is a fair starting point, not a rule.

Quick check

A projectile is at the highest point of its path. Which statement about its velocity there is correct?

Key points

Resolve the launch velocity: $u_x = u\cos\theta$ , $u_y = u\sin\theta$ ; then treat the two motions separately.
On level ground: $T = \frac{2u\sin\theta}{g}$ , $H = \frac{u^{2}\sin^{2}\theta}{2g}$ , $R = \frac{u^{2}\sin 2\theta}{g}$ ; the path $y = x\tan\theta - \frac{gx^{2}}{2u^{2}\cos^{2}\theta}$ is a parabola.
$45^\circ$ gives the greatest range $\frac{u^{2}}{g}$ ; complementary angles give equal ranges.
From a height, solve the quadratic for $t$ ; air resistance shortens the range and spoils the symmetry.

Sources and credits

Photo: Parabolic water trajectory by GuidoB, Wikimedia Commons, licensed under CC BY-SA 3.0.
Video: Projectile Motion: Maximum Range Angle, The Organic Chemistry Tutor, embedded from YouTube.

Written by Ganzaa: the lesson text, derivations, diagrams, graphs, worked examples and practice exercises. This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1The best angle

2Splitting an oblique launch

3Time of flight, maximum height and range

4The best angle, and pairs of angles

5Applications, and the effect of air

The best angle

Splitting an oblique launch

Time of flight, maximum height and range

The best angle, and pairs of angles

Applications, and the effect of air