11.05 Vectors and projectile motion
Projectile motion
The motion of a projectile generally makes a curved path. To understand such a trajectory, we consider the motion to be made up of a horizontal part and a vertical part. The combination of these two gives the observed path of the projectile.
The two components of the motion of a projectile are combined by means of the vector sum. For example, the horizontal component of velocity is constant in magnitude if the effect of air resistance is ignored. The vertical component of velocity changes with constant downward acceleration according to the influence of gravity.
In the diagram, the horizontal velocity vector is represented by an arrow of constant length, the vertical velocity vector (in red) varies in magnitude. Here, the situation is shown at three different times $t_1$t1, $t_2$t2 and $t_3$t3. The sum in each case varies in both length and direction, giving velocity vectors $v_1$v1, $v_2$v2 and $v_3$v3.
To make specific calculations to do with the displacement and velocity vectors of a projectile, we need the equations of motion in a straight line that are discussed in Physics courses. The equations given below use the symbols $t$t for time, $s$s for displacement, $v$v for velocity ($v_0$v0 is the initial velocity) and $a$a for acceleration. Acceleration is assumed to be constant and if it happens that velocity is constant then $s=vt.$s=vt.
$s=\frac{v_0+v}{2}t$s=v0+v2t for constant acceleration
$v=v_0+at$v=v0+at
$s=v_0t+\frac{1}{2}at^2$s=v0t+12at2
$v^2=v_0^2+2as$v2=v20+2as
[Note that since velocity is the rate of change of displacement the second of these equations should be the derivative of the third equation, which it is when acceleration is constant. Also, the fourth equation is obtained from the first two by eliminating the variable $t$t.]
Acceleration
Acceleration due to gravity is about $g=-9.8$g=−9.8m/s$^2$2. This means the velocity of a projectile increases in the downward direction.
The horizontal acceleration is usually negligible although for some specialized purposes it may be necessary to take into account the effect of friction with the air, which would give a negative acceleration. We ignore air resistance for the moment.
Clearly, vector addition of the horizontal and vertical components of acceleration gives a resultant acceleration with magnitude $g$g pointing directly downwards.
Velocity
We treat the horizontal and vertical components separately and add them at a time $t$t to discover the velocity vector at that moment.
In the diagram, a projectile has been launched at an angle $\theta$θ from the horizontal. The magnitude of its velocity is $|v|$|v|.
Initially, the vertical component of the velocity is $|v|\sin\theta$|v|sinθ and the horizontal component is $|v|\cos\theta$|v|cosθ.
At a time $t$t after the launch, the vertical component is $|v|\sin\theta-gt$|v|sinθ−gt, using the second of the equations given above. The horizontal component stays constant at $|v|\cos\theta$|v|cosθ.
The magnitude of the resultant velocity at time $t$t is then found using Pythagoras's theorem and the direction is found using the arctangent function.
Worked example
Question 1
A projectile is launched at an initial velocity of $200$200m/s at an angle of $30^\circ$30° from the horizontal. Find the velocity vector at $t=10$t=10 seconds.
The vertical component of the velocity at $t=10$t=10 is
| $|v|\sin\theta-gt$|v|sinθ−gt | $=$= | $200\sin30^\circ-9.8\times10$200sin30°−9.8×10 |
| $=$= | $100-98$100−98 | |
| $=$= | $2$2 m/s |
The horizontal component at $t=10$t=10 is $200\cos30^\circ=100\sqrt{3}$200cos30°=100√3.
By Pythagoras, the magnitude of the resultant velocity is $\sqrt{30000+4}\approx173.2m/s$√30000+4≈173.2m/s.
The tangent of the angle from the horizontal must be $\frac{2}{\frac{100}{\sqrt{3}}}=\frac{\sqrt{3}}{50}$2100√3=√350. So, this angle is $\arctan\frac{\sqrt{3}}{50}\approx2^\circ$arctan√350≈2°.
Displacement
As before, we treat the horizontal and vertical components separately before combining them by vector addition. The calculations for displacement are needed when we wish to determine the path through space of the projectile.
The horizontal component is just $|v|\cos\theta t$|v|cosθt where $|v|$|v| is the magnitude of the initial velocity.
The vertical component is $|v|\sin\theta t-\frac{1}{2}\times9.8t^2$|v|sinθt−12×9.8t2.
Worked example
Question 2
What is the position of the projectile in Example $1$1 at $t=10$t=10 seconds?
The horizontal displacement is $\frac{100}{\sqrt{3}}\times10=577$100√3×10=577 m.
The vertical displacement is $100\times10-\frac{9.8}{2}\times10^2=510$100×10−9.82×102=510 m.
In terms of a Cartesian coordinate system, we can associate the horizontal displacement with the $x$x-axis and the vertical displacement with the $y$y-axis. A plot of the points obtained at a succession of times $t$t corresponds to the path of the projectile through space. We have
$x=|v|\cos\theta t$x=|v|cosθt
$y=|v|\sin\theta t-\frac{1}{2}\times9.8t^2$y=|v|sinθt−12×9.8t2
From this pair of parametric equations, we can eliminate $t$t to obtain y as a function of $x$x.
From the first equation, we have $t=\frac{x}{|v|\cos\theta}$t=x|v|cosθ, and so, $t^2=\frac{x^2}{|v|^2\cos^2\theta}$t2=x2|v|2cos2θ.
On substitution into the second parametric equation, we have
$y=|v|\sin\theta\frac{x}{|v|\cos\theta}-4.9\frac{x^2}{|v|^2\cos^2\theta}$y=|v|sinθx|v|cosθ−4.9x2|v|2cos2θ
This is a quadratic function of $x$x and we conclude that the path of the projectile is parabolic in shape.
Question 3
How far will a projectile launched from a flat horizontal surface travel if its initial velocity has magnitude $|v|$|v| and it is launched at an angle of $\theta^\circ$θ°?
The time during which the projectile is in the air can be deduced using the vertical component of the displacement. The height above the ground is $h=|v|\sin\theta t-\frac{1}{2}\times9.8t^2$h=|v|sinθt−12×9.8t2. When the projectile reaches the ground again after its flight, we have $h=0$h=0. So, $|v|\sin\theta t-\frac{1}{2}\times9.8t^2=0$|v|sinθt−12×9.8t2=0, which can be solved for $t$t.
The solutions are $t=0$t=0, which corresponds to the moment of launch, and $t=\frac{|v|\sin\theta}{4.9}$t=|v|sinθ4.9 when it reaches the ground again.
We can substitute this value of $t$t into the equation for the horizontal displacement. Thus,
| $D$D | $=$= | $|v|\cos\theta t$|v|cosθt |
| $=$= | $|v|\cos\theta\frac{|v|\sin\theta}{4.9}$|v|cosθ|v|sinθ4.9 | |
| $=$= | $|v|^2\frac{2\sin\theta\cos\theta}{9.8}$|v|22sinθcosθ9.8 | |
| $=$= | $\frac{|v|^2}{9.8}\sin2\theta$|v|29.8sin2θ |
We see that the distance $D$D depends on the initial velocity vector. The maximum distance for a given magnitude of the velocity is attained when $\sin2\theta$sin2θ is greatest. This occurs when $2\theta=90^\circ$2θ=90° and so, $\theta=45^\circ$θ=45°.
Horizontal and vertical equations of projectile motion
The path of a projectile near the surface of the earth is curved due to the fact that in the vertical direction the force of gravity is acting on the projectile while in the horizontal direction there is almost no force acting.
We consider the equations of motion in the two orthogonal directions separately and combine the results as a vector sum.
In the vertical direction, there is a change in the velocity of the projectile due to gravity. This acceleration is approximately $-9.8$−9.8 m/s$^2$2, also known as $1$1 $g$g. It is taken to be constant although it does change very slightly from place to place due to differences in geology. Far from the center of gravity of the earth or in the vicinity of other planets, the acceleration due to gravity is different.
The negative sign is used in the equations of motion to indicate that the acceleration is downwards.
In the horizontal direction, there is a friction force acting against the direction of motion. Its effect is small compared with the gravitational force and so it is ignored for most purposes.
The following equations of motion for constant acceleration are relevant. The letters $u$u and $v$v are used for initial and final velocities respectively. Displacement is given the letter $s$s, $a$a stands for acceleration and $t$t is elapsed time.
$s=s_0+ut+\frac{1}{2}at^2$s=s0+ut+12at2
$s=s_0+\left(\frac{u+v}{2}\right)t$s=s0+(u+v2)t
$v=u+at$v=u+at
$v^2=u^2+2as$v2=u2+2as
Worked examples
Question 4
A stone is thrown with a velocity of $40$40 m/s at an angle of $30^\circ$30° from the horizontal. Describe the vertical and horizontal components of the motion.
Displacement
The vertical component of the initial velocity is $40\sin30^\circ=20$40sin30°=20 m/s.
The horizontal component is $40\cos30^\circ=20\sqrt{3}$40cos30°=20√3 m/s.
Substituting into the equation $s=s_0+ut+\frac{1}{2}at^2$s=s0+ut+12at2, we obtain the displacement vector
$\mathbf{s}=\left(20\sqrt{3}t,20t-4.9t^2\right)$s=(20√3t,20t−4.9t2)
(We let the origin be the point from which the stone was thrown so that $s_0=0$s0=0. The acceleration in the horizontal direction is $0$0 which means the $\frac{1}{2}at^2$12at2 term disappears in the horizontal component.)
The vertical component, $20t-4.9t^2$20t−4.9t2, is $0$0 initially but there is a second solution to this quadratic, which gives the time at which the stone returns to ground level. If
$20t-4.9t^2=0$20t−4.9t2=0
Either $t=0$t=0 or $t=20-4.9t=0\Rightarrow t=\frac{20}{4.9}\approx4.08$t=20−4.9t=0⇒t=204.9≈4.08s.
If $t=4.08$t=4.08 is substituted into the expression for the horizontal component of the displacement, we find that the stone traveled a horizontal distance of $20\sqrt{3}\times4.08\approx141.3$20√3×4.08≈141.3m.
To find the maximum height reached by the stone, we can use $v^2=u^2+2as$v2=u2+2as. At the greatest point $v=0$v=0. Thus, $0=20^2-2\times9.8s$0=202−2×9.8s. Solving for $s$s, we find $s\approx20.4$s≈20.4 m.
Velocity
Using $v=u+at$v=u+at and the initial velocity components, we deduce that the velocity vector is given by
| $\mathbf{v}$v | $=$= | $\left(20\sqrt{3}t,20-9.8t\right)$(20√3t,20−9.8t) |
| $=$= | $\left(34.6t,\ 20-9.8t\right)$(34.6t, 20−9.8t) |
Note that at $t=4.08$t=4.08 s, the velocity in the vertical direction is $-20$−20 m/s, the reverse of the initial velocity.
Question 5
Using the displacement vector from Question $4$4, we can deduce the shape of the trajectory of the stone.
We have $\mathbf{s}=\left(20\sqrt{3}t,20t-4.9t^2\right)$s=(20√3t,20t−4.9t2). We let the horizontal component be the values of the $x$x-axis and the vertical component be the values on the $y$y-axis. So,
$x=20\sqrt{3}t$x=20√3t and
$y=20t-4.9t^2$y=20t−4.9t2
If we can eliminate the parameter $t$t from these two equations, we will have a relation between $x$x and $y$y only.
From the first equation, $t=\frac{x}{20\sqrt{3}}$t=x20√3. This is substituted into the second equation to obtain
$y=20\times\frac{x}{20\sqrt{3}}-4.9\left(\frac{x}{20\sqrt{3}}\right)^2$y=20×x20√3−4.9(x20√3)2
and so, we have the quadratic $y\approx0.577x-0.004x^2$y≈0.577x−0.004x2 and the trajectory must be a parabola.
Magnitude and direction of the velocity of a projectile
We saw that the equation $v=u+at$v=u+at could be applied in the horizontal and vertical directions to obtain a velocity vector that varied with time $t$t. For example, we could have $\mathbf{v}(t)=\left(1.5,2-9.8t\right)$v(t)=(1.5,2−9.8t)m/s.
In this case, the horizontal velocity is constant at $1.5$1.5m/s and the vertical velocity varies according to the rule $v=2-9.8t$v=2−9.8t.
Magnitude
We use the Pythagorean theorem to calculate the magnitude or 'length' of a vector. So, we have
$\left|\mathbf{v}(t)\right|=\sqrt{1.5^2+\left(2-9.8t\right)^2}$|v(t)|=√1.52+(2−9.8t)2
This formula will give the magnitude of the velocity at any time $t$t for which the question makes sense.
We could write a more general formula using $u_x$ux and $u_z$uz for the initial velocities in the horizontal and vertical directions:
$\left|\mathbf{v}(t)\right|=\sqrt{u_x^2+\left(u_z-9.8t\right)^2}$|v(t)|=√u2x+(uz−9.8t)2
In three dimensions, two horizontal and one vertical, we would have a very similar equation:
$\left|\mathbf{v}(t)\right|=\sqrt{u_x^2+u_y^2+\left(u_z-9.8t\right)^2}$|v(t)|=√u2x+u2y+(uz−9.8t)2
Direction
Velocity is a vector quantity, so it has direction.
The direction of a vector can be given in terms of the cosines of the angles it makes with the axes. The diagram shows the two-dimensional case.
We see that the cosine of the angle with the horizontal axis is $\cos\alpha=\frac{u_x}{\left|\mathbf{v}\right|}$cosα=ux|v|.
The cosine of the angle with the vertical axis is $\cos\beta=\frac{u_z-9.8t}{\left|\mathbf{v}\right|}$cosβ=uz−9.8t|v|.
Notice that the two angles are not independent. That is, if we know one we can find the other, as follows.
| $\cos^2\alpha$cos2α | $=$= | $\frac{u_x^2}{\left|\mathbf{v}\right|^2}$u2x|v|2 |
| $\cos^2\beta$cos2β | $=$= | $\frac{\left(u_z-9.8t\right)^2}{\left|\mathbf{v}\right|^2}$(uz−9.8t)2|v|2 |
| $\therefore\ \ \cos^2\alpha+\cos^2\beta$∴ cos2α+cos2β | $=$= | $\frac{u_x^2+\left(u_z-9.8t\right)^2}{\left|\mathbf{v}\right|^2}$u2x+(uz−9.8t)2|v|2 |
| $=$= | $\frac{\left|\mathbf{v}\right|^2}{\left|\mathbf{v}\right|^2}$|v|2|v|2 | |
| $=$= | $1$1 |
Thus, $\cos^2\beta=1-\cos^2\alpha$cos2β=1−cos2α
In three dimensions, we would also have the angle $\gamma$γ made by $\mathbf{v}$v with the $y$y-axis. If the $y$y-component of the velocity is $u_y$uy then it can be shown that $\cos\gamma=\frac{u_y}{\left|\mathbf{v}\right|}$cosγ=uy|v|.
Once again, the sum of the squares of the three cosines is one but this time the angles do not necessarily lie in horizontal or vertical planes.
Worked examples
Question 6
Given a velocity vector function $\mathbf{v}(t)=\left(1.5,1,2-9.8t\right)$v(t)=(1.5,1,2−9.8t)m/s, find the magnitude and the three direction cosines at $t=0.2$t=0.2.
At $t=0.2$t=0.2, the velocity vector is given by $\left(1.5,1,0.04\right)$(1.5,1,0.04). Its magnitude is
$\sqrt{1.5^2+1^2+0.04^2}=1.8$√1.52+12+0.042=1.8m/s.
The direction cosines relative to the $x$x-, $y$y- and $z$z-axes respectively are
$\frac{1.5}{1.8}\approx0.8333$1.51.8≈0.8333
$\frac{1}{1.8}\approx0.5556$11.8≈0.5556
$\frac{0.04}{1.8}\approx0.0222$0.041.8≈0.0222
The angles themselves can be found easily from the cosines.
Question 7
Suppose a velocity vector has direction cosines $\sqrt{0.5}$√0.5, $\sqrt{0.3}$√0.3 and $\sqrt{0.2}$√0.2 at a time $t=15$t=15s. The magnitude of the velocity vector at that moment is $76$76m/s. Find the three components of the velocity vector.
For each component, we have $\cos\theta=\frac{u_i}{\left|\mathbf{v}\right|}$cosθ=ui|v|. So, the components are
$u_x=\sqrt{0.5}\times76$ux=√0.5×76
$u_y=\sqrt{0.3}\times76$uy=√0.3×76
$u_z=\sqrt{0.2}\times76$uz=√0.2×76
Hence, the velocity vector at $t=15$t=15s is $\left(53.7,41.6,34.0\right)$(53.7,41.6,34.0)m/s.