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1.4Projectile motion

Figure 1.32 shows the positions of two objects every 0.2 s: the first was simply allowed to drop vertically from rest; the other was launched horizontally with no vertical component of velocity. We see that in the vertical direction, both objects fall the same distance in the same time and so will hit the ground at the same time.

Figure 1.32: A body dropped from rest and one launched horizontally cover the same vertical distance in the same time.

theory of knowledge

How do we understand this fact? Consider Figure 1.33, in which a black ball is projected horizontally with velocity v. A blue ball is allowed to drop vertically from the same height.

Figure 1.33a shows the situation when the balls are released as seen by an observer X at rest on the ground. But suppose there is an observer Y, who moves to the right with velocity v 2 with respect to the ground. What does Y see? Observer Y sees the black ball moving to the right with velocity v 2 and the blue ball approaching with velocity −v 2 Figure 1.18b). The motions of the two balls are therefore identical (except for direction). So this observer will determine that the two bodies reach the ground at the same time. Since time is absolute in Newtonian physics, the two bodies must reach the ground at the same time as far as any other observer is concerned as well.

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Figure 1.33: a A ball projected horizontally and one simply dropped from rest from the point of view of observer X. Observer Y is moving to the right with velocity v 2 with respect to the ground. b From the point of view of observer Y, the black and the blue balls have identical motions.

The motion of a ball that is projected at some angle can be analysed by separately looking at the horizontal and the vertical directions. All we have to do is consider two motions, one in the horizontal direction in which there is no acceleration and another in the vertical direction in which we have an acceleration, g.

Consider Figure 1.34, where a projectile is launched at an angle θ to the horizontal with speed u.

The components of the initial velocity vector are u x = u cos θ and u y = u sin θ.

At some later time t the components of velocity are v x and v y. In the x-direction there is zero acceleration and in the y-direction, the acceleration is −g and so:

Horizontal direction | Vertical direction
v x = u cos θ | v y = u sin θ − gt

Figure 1.34: A projectile is launched at an angle θ to the horizontal with speed u.

The green vector in Figure 1.35a shows the position of the projectile t seconds after launch. The red arrows in Figure 1.35b show the velocity vectors and their components. Here the true vector nature of what we have been calling position comes into its own.

The position of the projectile is determined by the position vector the vector from the origin of a coordinate system to the position of a particle r→, a vector from the origin to the position of the projectile. The components of this vector are x and y.

Figure 1.35: a The position of the particle is determined if we know the x- and y-components of the position vector. b The velocity vectors for projectile motion are tangents to the parabolic path.

Exam tip

All that we are doing is using the formulas from the previous section for velocity and position – v = u + at and s=u t+1 2 a t 2 – and rewriting them separately for each direction x and y. In the x-direction there is zero acceleration, and in the y-direction there is an acceleration –g.

We would like to know the x- and y-components of the position vector. We now use the formula for position. In the x-direction:

x = u x t and so x = ut cos θ

In the y-direction: y=u y t−1 2 g t 2 and
so y=u t sin θ−1 2 g t 2.

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In summary:

Horizontal direction | Vertical direction
v x = u cos θ | v y = u sin θ − gt
x = ut cos θ | y=u t sin θ−1 2 g t 2

The equation with ‘squares of speeds’ is a bit trickier (carefully review the following steps). It is:

v 2=u 2−2 g y

Since v 2 = v x 2 + v y 2 and u 2 = u x 2 + u y 2, and in addition v x 2 = u x 2, this is also equivalent to:

v x 2=u y 2−2 g y

Exam tip

Always choose your x- and y-axes so that the origin is the point where the launch takes place.

CHECK yourself 5

A projectile is launched with kinetic energy K at 60° to the horizontal. What is the kinetic energy of the projectile at the highest point of its path? (Kinetic energy is equal to 1 2 m v 2.)

Worked Example 1.10

A body is launched with a speed of 18.0 m s−1 at the following angles:

a30° to the horizontal

b0° to the horizontal

c90° to the horizontal

Find the x- and y-components of the initial velocity in each case.

Answer

a v x=u cos θ v y=u sin θ v x=18.0×cos 30°v y=18.0×sin 30°v x=15.6 m s−1 v y=9.00 m s−1

b v x=18.0 m s−1 v y=0 m s−1

c v x=0 v y=18.0 m s−1

Worked Example 1.11

For a projectile launched at some angle above the horizontal, sketch graphs to show the variation with time of:

athe horizontal component of velocity

bthe vertical component of velocity

Answer

The graphs are shown in Figure 1.36.

Figure 1.36: Answer to worked example 1.11

Worked Example 1.12

An object is launched horizontally from a height of 20 m above the ground with speed 15 m s−1. Determine:

athe time at which it will hit the ground

bthe horizontal distance travelled

cthe speed with which it hits the ground

(Take g = 10 m s−2.)

Answer

aThe launch is horizontal, i.e. θ = 0°, and so the formula for vertical position is just y=−1 2 g t 2.

The object will hit the ground when y = −20 m.

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Substituting the values, we find:

−20=−5 t 2⇒t=2.0 s

bThe horizontal position is found from x = ut. Substituting values:

x = 15 × 2.0 = 30 m

(Remember that θ = 0°.)

cUse v 2 = u 2 − 2gy to get:

v 2=15 2−2×10×(−20)

v=25 m s−1

Worked Example 1.13

An object is launched horizontally with a velocity of 12 m s−1. Determine:

athe vertical component of velocity after 4.0 s

bthe x- and y-components of the position vector of the object after 4.0 s.

Answer

aThe launch is again horizontal, i.e. θ = 0°, so substitute this value in the formulas. The horizontal component of velocity is 12 m s−1 at all times.

From v y = −gt, the vertical component after 4.0 s is v y = −40 m s−1.

bThe coordinates after time t are:

x=u t y=−1 2 g t 2 x=12.0×4.0 y=−5×16 x=48 m and y=−80 m

Exam tip

Worked example 1.13 is a basic problem— you must know how to do this!

Figure 1.37 shows an object thrown at an angle of θ = 30° to the horizontal with initial speed 20 m s−1. The position of the object is shown every 0.2 s. Note how the dots get closer together as the object rises (the speed is decreasing) and how they move apart on the way down (the speed is increasing). It reaches a maximum height of 5.1 m and travels a horizontal distance of 35 m. The photo in Figure 1.38 shows an example of projectile motion.

Figure 1.37: A launch at of θ = 30° to the horizontal with initial speed 20 m s−1.

At what point in time does the vertical velocity component become zero? Setting v y = 0 we find:

0=u sin θ−g t⇒t=u sin θ g

Figure 1.38: A real example of projectile motion.

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The time when the vertical velocity becomes zero is, of course, the time when the object attains its maximum height. What is this height? Going back to the equation for the vertical component of position, we find that when:

t=u sin θ g

y is given by:

y max=u u sin θ g sin θ−1 2 g(u sin θ g)2 y max=u 2 sin 2⁡θ 2 g

What about the maximum displacement in the horizontal direction (called the range)? At this point y is zero. We can find the time by setting y = 0 in the formula for y but it is easier to notice that since the path is symmetrical the time to cover the range is double the time to get to the top, so t=2 u sin θ g.

Exam tip

You should not remember these formulas by heart. You should be able to derive them quickly.

Therefore the range is:

x=2 u 2 sin θ cos θ g

A bit of trigonometry allows us to rewrite this as:

x=u 2 sin 2 θ g

One of the identities in trigonometry is 2 sinθ cosθ = sin2θ.

The maximum value of sin 2θ is 1, and this happens when 2θ = 90° (θ = 45°). In other words, we obtain the maximum range with a launch angle of 45°. This equation also says that there are two different angles of launch that give the same range for the same initial speed. These two angles add up to a right angle. (Can you see why?)

Check yourself 6

On earth the maximum height and range of a projectile are H and R, respectively. The projectile is launched with the same velocity on a planet where the acceleration of free fall is 2g. What are the maximum height and range of the projectile on the planet?

Worked Example 1.14

A projectile is launched at 32.0° to the horizontal with initial speed 25.0 m s−1. Determine the maximum height reached. (Take g = 9.81 m s−2.)

Answer

The vertical velocity is given by v y = u sin θ − gt and becomes zero at the highest point. Thus:

t=u sin θ g

t=25.0×sin 32.0°9.81

t=1.35 s

Substituting in the formula for y, y=u t sin θ−1 2 g t 2, we get:

y=25×sin 32.0°×1.35−1 2×9.81×1.35 2

y=8.95 m

Equivalently, 0=(u sin θ)2−2 g y⇒y=(25.0×sin 32.0°)2 2×9.81=8.95 m

Worked Example 1.15

A projectile is launched horizontally from a height of 42 m above the ground. As it hits the ground, the velocity makes an angle of 55° to the horizontal. Find the initial velocity of launch. (Take g = 9.8 m s−2.)

Answer

The time it takes to hit the ground is found from y=−1 2 g t 2. (Here θ = 0° since the launch is horizontal.)

The ground is at y = −42 m and so:

−42=−1 2×9.8 t 2⇒t=2.928 s

Using v y = u sin θ − gt, when the projectile hits the ground:

v y=0−9.8×2.928

v y=−28.69 m s−1

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We know the angle the final velocity makes with the ground (Figure 1.39). Hence:

tan 55°=|v y v x|

⇒v x=28.69 tan 55°v x=20.09≈20 m s−1

Figure 1.39

Links

In projectile motion we have a single constant force whose direction is always vertical. This force is the weight of the body. When the body is projected at an angle to the vertical the result is motion along a path that is parabolic. There is one other instance where again a force that is constant in magnitude and direction acts on a body. This is the case of motion of an electric charge in a uniform electric field. So we expect that a charge projected at an angle in a region of uniform electric field will also result in motion along a parabolic path and what we have learned in this chapter will apply to that case as well. See Chapter 19.

Fluid resistance

The discussion of the previous sections has neglected air resistance forces. In general, whenever a body moves through a fluid (gas or liquid) it experiences a fluid resistance force a speed dependent force opposing the motion of a body through a fluid that is directed opposite to the velocity. Typically F = kv for low speeds and F = kv 2 for high speeds (where k is a constant). The magnitude of this force increases with increasing speed.

Imagine dropping a body of mass m from some height. Assume that the force of air resistance on this body is F = kv. Initially, the only force on the body is its weight, which accelerates it downwards. As the speed increases, the force of air resistance also increases. Eventually, this force will become equal to the weight and so the acceleration will become zero: the body will then move at constant speed, called terminal speed the constant speed attained when the resistance force becomes equal to the force pushing the body, v T. This speed can be found from:

m g=k v T

which leads to:

v T=m g k

Figure 1.40 shows how the speed and acceleration vary for motion with an air resistance force that is proportional to speed. The speed eventually becomes the terminal speed, and the acceleration becomes zero. The initial acceleration is g.

Figure 1.40: a The variation with time of a speed. b Acceleration in motion with an air resistance force proportional to speed.

The effect of air resistance forces on projectiles is very pronounced. Figure 1.41 shows the positions of a projectile with (red) and without (blue) air resistance forces. With air resistance forces the range and maximum height are smaller and the shape is no longer symmetrical.

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The projectile hits the ground at a steeper angle.

Figure 1.41: The effect of air resistance on projectile motion.

Worked Example 1.16

The force of air resistance in the motion described by Figure 1.40 is given by F = 0.653v.

Determine the mass of the projectile.

Answer

The terminal speed is 30 m s−1 and is given by v T=m g k. Hence:

m=0.653×30 9.8

m≈2.0 kg

nature of science

The simple and the complex

Careful observation of motion in the natural world led to the equations for motion with uniform acceleration along a straight line that we have used in this section. Thinking about what causes an object to move links to the idea of forces. However, although the material in this section is perhaps some of the ‘easiest’ material in your physics course, it does not enable you to understand the falling of a leaf off a tree. The falling leaf is complicated because it is acted upon by several forces: its weight, but also air resistance forces that constantly vary as the orientation and speed of the leaf change. In addition, there is wind to consider as well as the fact that turbulence in air greatly affects the motion of the leaf. So the physics of the falling leaf is far away from the physics of motion along a straight line at constant acceleration. But learning the principles of physics in a simpler context allows its application in more involved situations.

TEST YOUR UNDERSTANDING

29A ball rolls off a table with a horizontal speed of 2.0 m s−1. The table is 1.3 m high. Calculate how far from the table the ball will land.

30Two particles are on the same vertical line. They are thrown horizontally with the same speed, 4.0 m s−1, from heights of 4.0 m and 8.0 m.

aCalculate the distance that will separate the two objects when both land on the ground.

bThe particle at the 4.0 m height is now launched with horizontal speed u such that it lands at the same place as the particle launched from 8.0 m. Calculate u.

31For an object thrown at an angle of 40° to the horizontal at a speed of 20 m s−1, draw graphs of:

a horizontal velocity against time

b vertical velocity against time

c acceleration against time.

32Determine the maximum height reached by an object thrown with speed 24 m s−1 at 40° to the horizontal.

33An object is thrown with speed 20.0 m s−1 at an angle of 50° to the horizontal. Draw graphs to show the variation with time of:

athe horizontal position

bthe vertical position

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34A cruel man takes aim horizontally at a chimp that is hanging from the branch of a tree, as shown in Figure 1.42.

The chimp lets go of the branch as soon as the hunter pulls the trigger. Treating the chimp and the bullet as point particles, determine if the bullet will hit the chimp.

Figure 1.42

35A ball is launched from the surface of a planet. Air resistance and other frictional forces are neglected. The graph in Figure 1.43 shows the position of the ball every 0.20 s.

Figure 1.43

aUse this graph to determine:

ithe components of the initial velocity of the ball,

iithe angle to the horizontal the ball was launched at,

iiithe acceleration of free fall on this planet.

bMake a copy of the graph and draw two arrows to represent the velocity and the acceleration vectors of the ball at t = 1.0 s.

cThe ball is now launched under identical conditions from the surface of a different planet where the acceleration due to gravity is twice as large. Draw the path of the ball on your graph.

36A stone is thrown with a speed of 20.0 m s−1 at an angle of 48° to the horizontal from the edge of a cliff 60.0 m above the surface of the sea.

aCalculate the velocity with which the stone hits the sea.

bDiscuss qualitatively the effect of air resistance on your answer to a.

37 aState what is meant by terminal speed.

bA ball is dropped from rest. The force of air resistance on the ball is proportional to the ball’s speed. Explain why the ball will reach terminal speed.

38A projectile is launched with speed u at 45° to the horizontal. The projectile is at height h at two different times.

aShow that the horizontal distance separating those points is u g u 2−4 g h.

bDeduce, using the result in part a, the maximum height reached by this projectile.

39A projectile is launched with some speed at some angle to the horizontal. At 1.0 s and at 5.0 s the height of the projectile from the ground is the same. What is the maximum height reached by this projectile? (Take g = 10 m s–2.)