Projectiles
Courtesy www.physics.ubc.ca/.../p420_00/
darren/web/range/range.html
Two Dimensional Motion
Courtesy Physics Lessons.com
What’s a Projectile?
 An object moving in two
dimensions under the
influence of gravity
alone.
You Predict
 Two identical balls
leave the surface of a
table at the same time,
one essentially dropped
the other moving
horizontally with a good
speed. Which hits the
ground first?
Courtesy of www.mansfieldct.org/schools/
mms/staff/hand/drop.jpg
Galileo’s Analysis
 Horizontal and vertical motions can be
analyzed separately
 Ball accelerates downward with uniform
acceleration –g
 Ball moves horizontally with no acceleration
 So ball with horizontal velocity reaches
ground at same time as one merely dropped.
Courtesy Glenbrook South Physics, Tom Henderson
Equations
Horizontal
 vx = vx0
 x = x0 +vx0t
Vertical
 vy = vy0 –gt
 y = y0 + vy0t – 1/2gt2
 vy2 = vy02 -2gy
Assuming y positive up; ax = 0, ay = -g = - 9.80 m/s2
Problem Solving
 Read carefully
 Draw Diagram
 Choose xy coordinates and origin
 Analyze horizontal and vertical separately
 Resolve initial velocity into components
 List knowns and unknowns


Remember vx does not change
vy = 0 at top
Examples
Horizontal launch
 Student runs off 10m high cliff at 5 m/s and lands in
water. How far from base of cliff?
 y = -1/2gt2 (with y up +)
 time to fall t = (2y/-g)1/2 = (-20/-9.80) 1/2 = 1.43 s
 x = vx0t = 5 x 1.43 = 7.1 m
Extension
 How fast would the student have to run to
clear rocks 10m from the base of the same
10m high cliff?
 Again t = (2y/-g)1/2 = 1.43 s
 x = vx0 t
 vx0 = x/t = 10m / 1.43s = 7.0 m/s
Upwardly Launched Projectile With
Velocity v and Angle q
 vx0 = v0cosq
 vy0 = v0sinq
 Time in air = 2 vy0/g
 Range = R = vx0t = v0cosq x 2 vy0/g =
(2v02/g) sinqcosq = (v02 /g )sin2q
 This is Range Formula
v
0

Only for y = y0
Gunner’s version sin2q = Rg/v02
Trigonometric identity: 2 sinqcosq = sin2q
q
R
Questions
 What is the acceleration vector at maximum
height?
9.80 m/s2 downward at all times
 How would you find the velocity at a given
time?
Use kinematics equations for vx and vy , then
find v by sqrt sum of squares of them.
 How would you find the height at any time?
Use kinematics equation for y
 How does the speed at launch compare with
that just before impact?
Same
Longest Range
 What angle of launch gives the longest range
and why? Assume the projectile returns to
the height from which launched.
 45 degrees; must maximize sin2q; maximum
value of sine is one, happens for 900; then q =
450
 If a launch angle q gives a certain range, what
other angle will give the same range and
why?
 Hint: R goes as sinqcosq
Moving Launch Vehicle
 If ball is launched from moving cart, where
will it land?
Simulation
 Link to simulation
 Virtual Lab
A Punt
 Football kicked from 1.00 m above ground at
20.0 m/s at angle above horizontal q0 = 370 .
Find range.
 Can we use range formula?
 No. It doesn’t apply since y ≠ y0
 Let x0 = y0 = 0
y = -1.00 m
Vy0 = v0sinq = 20 x 0.6 = 12m/s
Note: Projectile
motion is parabolic
Punt, continued
 Find time of flight using y = y0 + vy0t – 1/2gt2
 -1.00m = 0 + (12.0m/s)t – (4.90 m/s2)t2
 (4.90 m/s2)t2 - (12.0m/s)t – (1.00m) = 0
 t = 2.53s or -0.081 s (impossible)
 x = vx0t = v0cos370 t= (16.0 m/s)(2.53s) = 40.5
m
Using Formulas
 Be sure the formula
applies to the situation
– that the problem lies
within its “range of
validity”
 Make sure you
understand what is
going on.