10.4 Projectile Motion

Greg Kelly, Hanford High School, Richland, Washington

Photo by Vickie Kelly, 2002

Fort Pulaski, GA

One early use of calculus was to study projectile motion.

In this section we assume ideal projectile motion:

Constant force of gravity in a downward direction

Flat surface

No air resistance (usually)

We assume that the projectile is launched from the origin at time t =0 with initial velocity vo.

ov

Let o ov v

then cos sino o ov v v i j

The initial position is: r 0 0 0o i j

Horizontal component Vertical component

At the origin

Note: We will see this again

ov

Newton’s second law of motion:

Vertical acceleration

f ma2

2f

d rm

dt

Force = Mass (Acceleration)

Second derivative = acceleration

ov

Newton’s second law of motion:

The force of gravity is:

Force is negativeBecause gravity pulls downward

f ma f mg j2

2f

d rm

dt

ov

Newton’s second law of motion:

The force of gravity is:

f ma f mg j2

2f

d rm

dt

mg j2

2

d rm

dt

Substituting for f:

ov

Newton’s second law of motion:

The force of gravity is:

f ma f mg j2

2f

d rm

dt

mg j2

2

d rm

dt

And simplifying

2

2

d rg

dt j

Initial conditions: r r v when o o

drt o

dt

2o

1r v r

2 ogt t j

21r

2gt j 0 cos sin o ov t v t i j

o vdr

gtdt

j

We get:

“ + c ”

Because it’s our initial condition

21r cos sin 0

2 o ogt v t v t j i j

21r cos sin

2o ov t v t gt

i j

Vector equation for ideal projectile motion:

21r cos sin 0

2 o ogt v t v t j i j

21r cos sin

2o ov t v t gt

i j

Vector equation for ideal projectile motion:

Parametric equations for ideal projectile motion:

21cos sin

2o ox v t y v t gt

Rearranged

Horizontal component Vertical component

Example 1:A projectile is fired at 60o and 500 m/sec.Where will it be 10 seconds later?

500 cos 60 10x

2500x

21500sin 60 10 9.8 10

2y

3840.13y

Note: The speed of sound is 331.29 meters/secOr 741.1 miles/hr at sea level.

The projectile will be 2.5 kilometers downrange and at an altitude of 3.84 kilometers.

ov

21cos sin

2o ox v t y v t gt Parametric Equations

meters metersHorizontal component Vertical component

Now, the maximum height of a projectile occurs when the vertical velocity equals zero.

sin 0o

dyv gt

dt

sinov gt

sinovt

g

time at maximum height

This makes sense because the path of a projectile is a parabola so the maximum would occur at the vertex

Because object is not going up anymore at this point

Recall:

Then,

Velocity

The maximum height of a projectile occurs when the vertical velocity equals zero.

sin 0o

dyv gt

dt

sinov gt

sinovt

g

We can substitute this expression into the formula for height to get the maximum height.

2

max

sin sin1sin

2o o

o

v vy v g

g g

21sin

2oy v t gt

2 2

max

sin sin

2o ov v

yg g

This is the height because it is the vertical component

Simplifying, we get,

2

max

sin sin1sin

2o o

o

v vy v g

g g

21sin

2oy v t gt

2 2

max

si2

2

n sin

2o ov v

yg g

2

max

sin

2ov

yg

maximum

height

And multiplying by “1”,

we get:

210 sin

2ov t gt When the height is zero:

10 sin

2ot v gt

0t 1) The time at launch:

Vertical component

Now, if we factor out t we have:

For this to equal 0, two things can happen:

OR

210 sin

2ov t gt When the height is zero:

10 sin

2ot v gt

1sin 0

2ov gt

1sin

2ov gt

2 sinovt

g

time at impact

(flight time)

Which gives us the

If we take the expression for flight time (time at impact) and substitute it into the equation for x, we can find the range.

cos ox v t

cos 2 sin

oov

x vg

If we take the expression for flight time and substitute it into the equation for x, we can find the range.

cos ox v t

2 sincos o

o

vx v

g

2

2cos sinovx

g

2

sin 2ovx

g Range

(Simplifying)

The range is maximum when

2

sin 2ovx

g Range

sin 2 is maximum.

sin 2 1

2 90o 45o

Range is maximum when the launch angle is 45o.

Recall:

If we start with the parametric equations for projectile motion, we can eliminate t to get y as a function of x.

cosox v t

coso

xt

v

21sin

2oy v t gt

If we start with the parametric equations for projectile motion, we can eliminate t to get y as a function of x.

cosox v t

coso

xt

v

21sin

2oy v t gt

2

1sin

2cos cosoo o

x x

vy g

vv

This simplifies to:

22 2

tan2 coso

gy x x

v

which is the equation of a parabola since it is a quadratic function.

If we start somewhere besides the origin, the equations become:

coso ox x v t 21sin

2o oy y v t gt

21cos sin

2o ox v t y v t gt As opposed to the parametric equations for ideal projectile motion:

Example 4:A baseball is hit from 3 feet above the ground with an initial velocity of 152 ft/sec at an angle of 20o from the horizontal. A gust of wind adds a component of -8.8 ft/sec in the horizontal direction to the initial velocity.

The parametric equations become:

152cos 20 8.8ox t 23 152sin 20 16oy t t

ov oy 1

2g

21cos sin

2o ox v t y v t gt 21sin

2o oy y v t gt

These equations can be graphed on the TI-89 to model the path of the ball:

Note that the calculator is in degrees.

t2

Max height about 45 ft

Distance traveled about 442 ft

Timeabout

3.3 sec

Usingthe

tracefunction:

In real life, there are other forces on the object. The most obvious is air resistance.

If the drag due to air resistance is proportional to the velocity:

dragF kv (Drag is in the opposite direction as velocity.)

Equations for the motion of a projectile with linear drag force are given on page 546.

You are not responsible for memorizing these formulas.