Projectile Motion

Projectile motion is the motion of an object that has been given an initial velocity and then moves under gravity. In the standard introductory model, air resistance is neglected, The Gravitational Field is uniform, Earth is treated as locally flat, and the motion is described in an inertial frame fixed near Earth’s surface. Under these assumptions the projectile has constant downward acceleration

(1)

where g ≃ 9.81 m∕s² near Earth’s surface.

The central idea is that two-dimensional projectile motion is a constant acceleration problem whose horizontal and vertical components can be solved separately. The components share the same time variable, so they combine to form one curved trajectory.

1 The ideal projectile model

Choose the x axis horizontal and the y axis vertically upward. Let the projectile begin at

(2)

with initial speed v₀ at angle 𝜃₀ above the horizontal. Then

(3)

Thus

(4)

and

(5)

In the ideal model,

(6)

These two equations contain the basic physics of ideal projectile motion.

2 Independent horizontal and vertical motion

Because the horizontal acceleration is zero,

(7)

and

(8)

The vertical component is ordinary one-dimensional constant acceleration motion:

(9)

and

(10)

The horizontal and vertical equations are independent except that they describe the same object at the same time t.

3 Velocity during flight

At any time,

(11)

so

(12)

The speed is

(13)

The instantaneous direction of motion may be described by an angle ϕ measured from the positive horizontal direction:

(14)

At the apex of the trajectory,

(15)

but in general

(16)

Therefore the projectile is still moving at the top of its path. Its acceleration also remains −ge_y there.

4 The trajectory is a parabola

The parametric equations are

(17)

and

(18)

From the horizontal equation,

(19)

Substituting into the vertical equation gives

(20)

This is quadratic in x, so the ideal trajectory is a parabola.

The parabolic result depends on the assumptions of constant downward gravity and zero drag. Real long range trajectories need not be parabolic.

5 Time to the apex

The projectile reaches its highest point when

(21)

Therefore

(22)

which gives

(23)

The maximum height above the launch point follows from the vertical no time equation

(24)

At the apex v_y = 0, so

(25)

6 Same height launch and landing

If the projectile lands at the same vertical level from which it was launched, then

(26)

The vertical displacement equation becomes

(27)

Besides the trivial root T = 0, the nonzero flight time is

(28)

The horizontal range is

(29)

hence

(30)

Using

(31)

we obtain

(32)

7 Maximum range and complementary angles

For fixed v₀ and same height launch and landing,

(33)

The maximum possible value of sin 2𝜃₀ is one. Therefore the maximum ideal range occurs when

(34)

or

(35)

Also,

(36)

Thus complementary launch angles 𝜃₀ and 90^∘− 𝜃 ₀ give the same ideal same height range. The high angle trajectory has a longer flight time and greater maximum height.

These results are not generally true when launch and landing heights differ or when drag is important.

8 Unequal launch and landing heights

If

(37)

one should usually avoid forcing the same height formulas onto the problem. Instead solve the vertical equation directly:

(38)

Rearranging,

(39)

The quadratic formula gives

(40)

The physically relevant root depends on the problem. A positive root corresponding to the later intersection with the landing height is normally selected.

Once the flight time is known,

(41)

9 Horizontal launch as a special case

For a horizontal launch,

(42)

so

(43)

If the projectile falls through a vertical distance h,

(44)

which gives

(45)

The horizontal distance is then

(46)

This example makes the component independence especially clear: the fall time depends only on the vertical motion, while the horizontal speed determines how far the projectile travels during that time.

10 Speed as a function of height

The component equations also give a useful relation between speed and vertical position. Since

(47)

and

(48)

adding yields

(49)

Thus in the ideal model the speed at a given height depends only on that height and the initial speed, not on whether the projectile is rising or falling. In particular, if the projectile later returns to its launch height,

(50)

The velocity is not the same vector, because its vertical component has changed sign.

11 Choosing a sign convention

The common choice is +x horizontal in the launch direction and +y upward. Then

(51)

A different convention is allowed, but the signs in every equation must remain consistent. Many projectile motion errors come from inserting g as a positive number while simultaneously treating downward acceleration as if it were positive in an equation written for an upward positive axis.

A reliable procedure is:

1. draw the axes;
2. resolve the initial velocity into components;
3. write a_x and a_y with signs;
4. solve the vertical and horizontal equations separately;
5. use the common time to connect them;
6. check whether the result is physically consistent.

12 Model limitations

The ideal model is extremely useful, but its assumptions must be stated. It neglects air resistance and lift, treats g as constant, ignores Earth curvature and rotation, and assumes the local ground frame is sufficiently inertial for the problem.

For a thrown ball over tens of meters these approximations may be excellent. For long range artillery, rockets, atmospheric flight, or orbital trajectories they may fail badly. Those problems require later topics such as drag, rotating frames, variable gravity, and orbital mechanics.

13 Worked example 1: equal height launch

A projectile is launched from level ground with

(52)

at

(53)

Find the horizontal and vertical initial velocity components, time of flight, range, and maximum height. Use g = 9.81 m∕s².

The initial components are

(54)

and

(55)

Because launch and landing heights are equal,

(56)

The range is

(57)

The maximum height above launch is

(58)

14 Worked example 2: horizontal launch from a cliff

A ball leaves a horizontal cliff with speed

(59)

from a height

(60)

Find the flight time, horizontal range, impact speed, and impact angle below the horizontal.

Because v_0y = 0,

(61)

The range is

(62)

At impact,

(63)

and

(64)

Therefore

(65)

The impact angle below horizontal satisfies

(66)

so

(67)

15 Worked example 3: angled launch from an elevated point

A projectile is launched from a platform 8.0 m above the ground with speed

(68)

at

(69)

Find the time to hit the ground and the horizontal range.

The components are

(70)

and

(71)

Take ground level as y = 0. Then

(72)

Solving the quadratic and selecting the positive physical root gives

(73)

Therefore

(74)

The same height formula T = 2v₀ sin 𝜃₀∕g would be wrong here because the projectile lands below its launch height.

16 Worked example 4: two angles for the same range

A projectile is launched and lands at the same height. Its launch speed is

(75)

and the required range is

(76)

Find the two possible launch angles in the ideal model.

Use

(77)

Then

(78)

The two angles between 0^∘ and 180^∘ having this sine are

(79)

and

(80)

Thus

(81)

or

(82)

The two angles are complementary, as expected. The low angle flight takes about 1.70 s, while the high angle flight takes about 4.81 s.

17 Practice problems

1. A projectile is launched at 16 m∕s at 30^∘. Find v _0x and v_0y.
2. A projectile has v_0x = 12 m∕s and v_0y = 9 m∕s. Find its initial speed and launch angle.
3. A projectile is launched horizontally at 15 m∕s from a 20 m high platform. Find the flight time and horizontal range.
4. A projectile is launched from level ground at 22 m∕s and 40^∘. Find the time to the apex and maximum height.
5. For the projectile in Problem 4, find the total same height flight time and range.
6. A projectile is launched at 30 m∕s at 60^∘. Find its horizontal and vertical velocity components after 2.0 s.
7. A projectile is launched from y₀ = 5.0 m at 14 m∕s and 25^∘. Write the horizontal and vertical position functions using x₀ = 0.
8. A projectile launched from level ground has v₀ = 20 m∕s. Find the ideal same height range for 30^∘ and for 60^∘. Compare the results.
9. A projectile is launched from level ground at 18 m∕s and 45^∘. What is its speed at the apex?
10. A projectile is launched at 25 m∕s from a point 10 m above the landing level at 20^∘. Find the physical flight time and horizontal range.
11. A projectile moves through a point 4.0 m above its launch height. If its initial speed was 15 m∕s and drag is neglected, find its speed at that height.
12. Explain why 45^∘ does not necessarily maximize horizontal distance when the projectile lands at a different height from the launch point.

18 Answer check

1. v_0x = 13.9 m∕s, v_0y = 8.0 m∕s.
2. v₀ = 15.0 m∕s, 𝜃₀ = 36.9^∘.
3. t = 2.02 s, R = 30.3 m.
4. t_top = 1.44 s, H = 10.2 m.
5. T = 2.88 s, R = 48.6 m.
6. v_x = 15.0 m∕s, v_y = 6.36 m∕s.
7. x(t) = 12.69t m, y(t) = 5.0 + 5.92t − 4.905t² m.
8. Both ranges are approximately 35.3 m.
9. v = 18 cos 45^∘ = 12.7 m∕s.
10. t ≃ 2.54 s, R ≃ 59.8 m.
11. v = ≃ 12.1 m∕s.
12. Because the same height range formula R = v₀² sin 2𝜃∕g no longer applies; the flight time depends on the unequal vertical displacement.

19 Summary

Ideal projectile motion is constant acceleration motion in two dimensions. The horizontal component has zero acceleration while the vertical component has acceleration −g. Resolving the launch velocity into components gives

(83)

and

(84)

The parabolic trajectory, maximum height, time of flight, and range all follow from these equations. Same height shortcuts are useful but should only be used when their assumptions are satisfied. For unequal heights, the robust method is to solve the vertical equation for time and then use the horizontal motion.

References