Projectile Motion-Starter

What is the path that the bike and the water take called?

Projectile MotionA ball dropped from rest looks like this……..

Projectile MotionA ball rolled off the table at 10m/s with no gravity looks like this…………….

Put them both together………

Regardless of the air resistance, the vertical and the horizontal components of velocity of an object in projectile motion are independent.

Slo

win

g do

wn

in +y

dir. S

peeding up in -y dir.

Constant speed in +x dir. ax = 0

a y =

-g

ay =

-g

Projectile Motion

When the magnitude of the velocity is given and its direction

specified then its componentscan be computed easily

x

y

V

VX

VYVX = Vcosq

VY = Vsinq

You must use the polar angle in these formulas.

Example: Find the x and y components of the initial velocity vector shown if v = 10 and q = 60 degrees.

vix = vi cos q = 10 cos(60) = 5.00 m/s

viy = vi sin q = 10sin(60) = 8.66 m/s

Vi = 10 q = 60o

Another ViewThe projectile falls away from a straight line it would have takenif there were no gravity by the same distance it falls from rest.

Projectile Motion Equations

Vertical Direction

Dy = viyt - 5t2

vfy = viy - 10t

Horizontal Direction

Dx = vixt

vfx = vix

Note: We are using g = 10

Auxiliary Equations vix = vi cosq viy = vi sin q

Case 1 : Fired Horizontally

𝑇𝑖𝑚𝑒𝑖𝑛 h𝑡 𝑒 𝐴𝑖𝑟 𝑡=√ h5 𝑅𝑎𝑛𝑔𝑒𝑅=𝑣 𝑖𝑡

Case 1 : Example

= = 2 seconds.

= 50(2) = 100 meters

A cannon is fired horizontallyat 50m/s from a cliff 20m tall.

1. How long is it in the air?

2. What is the horizontal range.

Case 2 :Ground-to-Ground Motion

Time of Flight, T = (2vi sin )q /g

Range , R= vi 2sin (2 )q / g

Maximum Height H, H = (vi sin q )2 /2g

Case 2: ExampleThe muzzle velocity of a Howitzer is 563 m/s.If it is elevated at 60 degrees, what is its range and time of flight?

Range , R= vi 2sin (2 ) q = (563)2sin(120)/10

= 27450m = 27km( Due to air resistance, the actual range is less than this.)

Time of Flight, T = (2vi sinq) /g = 2(563) sin(60)/10 = 97.5 seconds

The Trajectory Equation

(1) yf = vi(sinq) – (g/2)t2

(2) xf = vi(cosq )t

To get the path or trajectory equation for a projectile, we need to find y as a function of x. Starting at (0,0):

If you solve (2) for t and plug it into (1) you get:

𝑦=𝑥𝑡𝑎𝑛𝜃− 𝑥2𝑔2¿¿

Trajectory

𝑦=𝑥𝑡𝑎𝑛𝜃− 𝑥2𝑔2¿¿

This has the form y = ax +bx2

, a parabola.

0 1 2 3 4 5 6 7 8 90

0.5

1

1.5

2

2.5

3

3.5

4

Projectile Path

x

y

Example: Find the equation of the parabola if v = 10 and q = 60 degrees.

2

ExitThe path of a projectile is given by: y = 2x –x2

What is the horizontal range? Hint: for what values of x is y = 0?