Honors/AP Physics

Mr. Velazquez

Projectile motion describes the motion of objects moving

through the air in two dimensions near the surface of the Earth.

Golf ball, thrown or batted baseball, basketball shot, kicked football

Speeding bullet, rocket, fired cannonball

For analysis, we focus only on the motion of the projectile after it has been fired or thrown and before it lands or is caught, while it is under the action of gravity alone.

Again, we will ignore air resistance for most of these cases.

Motion in Two Dimensions

The best way to understand the

motion of a projectile is by analyzing the horizontal and vertical components of its motion separately. Galileo was the first to describe this

phenomenon accurately

Vertically, the object is experiencing a downward acceleration due to gravity (regardless of it’s horizontal motion).

Horizontally, the object maintains whatever horizontal velocity it had when it was launched, since it is not experiencing any horizontal acceleration.

Motion of a Projectile

Conceptual Question: Two balls having different speeds roll off the edge of a horizontal table at the same time.

Which hits the floor sooner, the faster ball or the slower one?

Motion of a Projectile

For a ball rolled off a table with initial velocity 𝑣0 in the

horizontal direction:

Horizontally speaking, the ball will maintain a velocity of 𝑣0 (which we may call 𝑣𝑥0) throughout its entire trajectory, until it hits the ground, which means 𝑣𝑥 final = 𝑣𝑥0

Vertically speaking, the ball will leave the edge of the table with no vertical velocity (𝑣𝑦0 = 0), but because it is being

acted upon by gravity in a downward direction, it will begin accelerating downward at a rate of 𝑔 = 9.807 𝑚 𝑠2

and will hit the ground with some final vertical velocity, 𝑣𝑦

Both final values for 𝑣𝑥 and 𝑣𝑦 can be combined to form a

single velocity vector with a magnitude and direction.

Motion of a Projectile

Red ball: rolled off the edge of a table; experiencing projectile motion Blue ball: dropped from the edge of a table Green ball: rolling across top of table with constant horizontal velocity equal to red ball

Basic Projectile Example

For a ball thrown or launched from the ground with

initial velocity 𝑣0 applied at some upward angle 𝜃0, the analysis will be similar: Horizontally speaking, the ball will maintain a velocity of

𝑣𝑥0 throughout its entire trajectory, until it hits the ground, which still means 𝑣𝑥 final = 𝑣𝑥0

Vertically speaking, now that there is an initial vertical component of the velocity, the ball will begin to decelerate at a rate of 𝑔 = −9.807 𝑚 𝑠2 the moment it is thrown, because it is still being acted upon by gravity in a downward direction; it will reach a maximum height (where 𝑣𝑦 = 0), then begin accelerating downward and will hit the ground with some final vertical velocity, 𝑣𝑦

Both final values for 𝑣𝑥 and 𝑣𝑦 can be combined to form a single velocity vector with a magnitude and direction.

Motion of a Projectile

Component velocities of a projectile

Projectile Fired at an Angle

Horizontal Motion (𝒂𝒙 = 𝟎, 𝒗𝒙 = constant)

Vertical Motion (𝒂𝒚 = −𝒈 = constant)

𝑣𝑥 = 𝑣𝑥0 𝑣𝑦 = 𝑣𝑦0 − 𝑔𝑡

𝑥 = 𝑥0 + 𝑣𝑥0𝑡 ∆𝑥 = 𝑣𝑥0𝑡

𝑦 = 𝑦0 + 𝑣𝑦0𝑡 −1

2𝑔𝑡2

∆𝑦 = 𝑣𝑦0𝑡 −1

2𝑔𝑡2

𝑣𝑦2 = 𝑣𝑦0

2 − 2𝑔∆𝑦

Kinematic Equations for Projectile Motion

Assumptions: y is positive upward, 𝑎𝑥 = 0, 𝑎𝑦 = −𝑔 = −9.807 𝑚 𝑠2

Note: 𝑔 is already assumed negative in these formulas, so leave the negative sign out when plugging it in.

If we assume that velocity 𝑣0 is applied a positive angle of 𝜃0 above the horizontal axis, we can also use the following equations to resolve this velocity into its component vectors:

𝒗𝒙𝟎 = 𝒗𝟎 𝐜𝐨𝐬𝜽𝟎

𝒗𝒚𝟎 = 𝒗𝟎 𝐬𝐢𝐧𝜽𝟎

Equations for Velocity Vectors

We can now see that the path of a projectile follows a parabolic arc, and we can even derive an equation describing its motion in terms of y and x. We set 𝑥0 = 𝑦0 = 0:

Parabolic Projectile Motion

𝑥 = 𝑣𝑥0𝑡

𝑦 = 𝑣𝑦0𝑡 −1

2𝑔𝑡2

𝑡 = 𝑥𝑣𝑥0

𝑦 = 𝑣𝑦0𝑥𝑣𝑥0 −

1

2𝑔 𝑥

𝑣𝑥0 2

𝒚 =𝒗𝒚𝟎

𝒗𝒙𝟎𝒙 −

𝒈

𝟐𝒗𝒙𝟎𝟐

𝒙𝟐

Substituting 𝑣𝑥0 = 𝑣0 cos 𝜃 and 𝑣𝑦0 = 𝑣0 sin 𝜃:

𝒚 = 𝐭𝐚𝐧𝜽𝟎 𝒙 −𝒈

𝟐𝒗𝟎𝟐 𝐜𝐨𝐬𝟐 𝜽𝟎

𝒙𝟐

1. Read the problem carefully and choose the object

2. Draw a diagram

3. Choose an origin and an xy coordinate system

4. Decide on a time interval

1. Should only include motion while under the effect of gravity alone

2. Will be the same interval for both x and y analyses

5. Examine the horizontal (x) and vertical (y) motions separately

1. If given an initial velocity, resolve it into x and y components

6. List known and unknown quantities

1. Include 𝑎𝑥 = 0, 𝑎𝑦 = −𝑔 or +𝑔 depending on whether you choose y positive up or down

7. Apply relevant equations

1. Combine equations if necessary

Solving Projectile Problems

A stunt driver on a motorcycle speeds horizontally off a 50.0-m high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are?

Example Problem

A football is kicked at an angle 𝜃 = 37.0° above level ground with a velocity of 20.0 m/s. Calculate:

a) The maximum height of the ball

b) The time of travel before the football hits the ground

c) How far away it hits the ground

Example Problem

Derive a formula for the horizontal range, R, of a projectile in terms of its initial velocity and angle. The horizontal range is defined as the horizontal distance the projectile travels before returning to its original height (typically on the ground), or 𝑦 (final)= 𝑦0. In other words, find an equation that would give us the horizontal distance a projectile will travel (R), if we are given its initial velocity vector (𝑣0 and 𝜃).

Example Problem

A cannonball is fired with an initial speed of 65.2 m/s at an angle 34.5° above the horizontal, on a long, flat firing range.

(a) Sketch the trajectory of the cannonball

(b) Find the maximum height reached by the cannonball

(c) Find its total time spent in the air

(d) Find the total horizontal distance covered

(e) Calculate the velocity of the projectile 1.50 s after it’s fired

Classwork/Exit Ticket