10.4 Projectile Motion

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

ov

Newton’s second law of motion:

Vertical acceleration

f ma2

2f

d rm

dt

ov

Newton’s second law of motion:

The force of gravity is:

Force is in the downward direction

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

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

2

2

d rg

dt j

Initial conditions:

21r cos sin

2o ov t v t gt

i j

Vector equation for ideal projectile motion:

Vector equation for ideal projectile motion:

Parametric equations for ideal projectile motion:

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

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

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

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

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

21sin

2oy v t gt

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

210 sin

2ov t gt When the height is zero:

10 sin

2ot v gt

0t time at launch:

210 sin

2ov t gt When the height is zero:

10 sin

2ot v gt

0t time at launch:1

sin 02ov gt

1sin

2ov gt

2 sinovt

g

time at impact

(flight time)

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

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

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.

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.

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

coso ox x v t 21sin

2o oy y v t gt

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

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.