More Projectile Motion Discussion:
Examples
More Projectile Motion Discussion:
Examples
I hope this doesn’t
apply to you!
Solving Projectile Motion Problems
1. Read the problem carefully, & choose the object(s) you
are going to analyze.
2. Sketch a diagram.
3. Choose an origin & a coordinate system.
4. Decide on the time interval; this is the same in both
directions, & includes only the time the object is
moving with constant acceleration g.
5. Solve for the x and y motions separately.
6. List known & unknown quantities. Remember that vx
never changes, & that vy = 0 at the highest point.
7. Plan how you will proceed. Use the appropriate
equations; you may have to combine some of them.
Example 4.4: Non-Symmetric Projectile Motion
A stone is thrown! xi = yi = 0
yf = -45.0 m, vi = 20 m/s, θi = 30º
a) Time to hit the ground?
b) Speed just before it hits?
c) Distance from the base of the
building where it lands?
Kinematic Equations
vxi = vicosθi, vyi = visinθi
vxf = vxi , xf = vxi t
vyf = vyi - gt
yf = vyi t - (½)gt2
(vyf) 2 = (vyi)2 - 2gyf
Example 4.4: Solution
A stone is thrown! xi = yi = 0
yf = -45.0 m, vi = 20 m/s, θi = 30º
a) Time to hit the ground?
b) Speed just before it hits?
c) Distance from the base of the
building where it lands?
First, calculate
vxi = vi cos(θi) = 17.3 m/s
vyi = vi sin(θi) = 10.0 m/s
a) Time to hit the ground?
(Time when yf = -45.0 m)
yf = -45m = vyi t - (½)gt2
A general quadratic must be solved
using the quadratic equation! This
gives:
t = 4.22 s
Kinematic Equations
vxi = vicosθi, vyi = visinθi
vxf = vxi , xf = vxi t
vyf = vyi - gt
yf = vyi t - (½)gt2
(vyf) 2 = (vyi)2 - 2gyf
Example 4.4: Solution
A stone is thrown! xi = yi = 0
yf = -45.0 m, vi = 20 m/s, θi = 30º
a) Time to hit the ground?
b) Speed just before it hits?
c) Distance from the base of the
building where it lands?
First, calculate
vxi = vi cos(θi) = 17.3 m/s
vyi = vi sin(θi) = 10.0 m/s
thit = 4.22 s
b) Velocity just before it hits?
vxf = vxi , vyf = vyi – gt so vxf = 17.3 m/s
vyf = 10 – (9.8)(4.22) = - 31.3 m/s
Speed (vf)2 = (vxf)2 + (vyf)2
vf = 35.8 m/s
Angle: tan(θf) = (vyf/vxf) = -(31.3/17.3) = -1.8
θf = -60.9º
Kinematic Equations
vxi = vicosθi, vyi = visinθi
vxf = vxi , xf = vxi t
vyf = vyi - gt
yf = vyi t - (½)gt2
(vyf) 2 = (vyi)2 - 2gyf
Example 4.4: Solution
A stone is thrown! xi = yi = 0
yf = -45.0 m, vi = 20 m/s, θi = 30º
a) Time to hit the ground?
b) Speed just before it hits?
c) Distance from the base of the
building where it lands?
First, calculate
vxi = vi cos(θi) = 17.3 m/s
vyi = vi sin(θi) = 10.0 m/s
thit = 4.22 s
vf = 35.8 m/s, θf = -60.9º
c) Distance from the base of th
building where it lands?
xf = vxi thit = (17.3)(4.22) = 73.0 m
Kinematic Equations
vxi = vicosθi, vyi = visinθi
vxf = vxi , xf = vxi t
vyf = vyi - gt
yf = vyi t - (½)gt2
(vyf) 2 = (vyi)2 - 2gyf
Example 4.2: The Long Jump
A long-jumper leaves the ground at an angle θi = 20° above the
horizontal at a speed of vi = 8.0 m/s.
a) How far does he jump in the horizontal direction?
(Assume his motion is equivalent to that of a particle.)
b) What is the maximum height reached?
Kinematic Equations
vxi = vicosθi, vyi = visinθi ,vxf = vxi
xf = vxi t, vyf = vyi – gt
yf = vyi t - (½)gt2
(vyf) 2 = (vyi)2 - 2gyf
The Long Jump: Solutions
A long-jumper leaves the ground at an angle θi = 20° above the
horizontal at a speed of vi = 8.0 m/s.
a) How far does he jump in the horizontal direction?
(Assume his motion is equivalent to that of a particle.)
b) What is the maximum height reached?
vxi = vi cos(θi) = 7.5 m/s
vyi = vi sin(θi) = 4.0 m/s
a) How far does he jump in the
horizontal direction? Range =
R = (2vxivyi/g) = 2(7.5)(4)/(9.8)
R = 7.94 m
Kinematic Equations
vxi = vicosθi, vyi = visinθi ,vxf = vxi
xf = vxi t, vyf = vyi – gt
yf = vyi t - (½)gt2
(vyf) 2 = (vyi)2 - 2gyf
The Long Jump: Solutions
A long-jumper leaves the ground at an angle θi = 20° above the
horizontal at a speed of vi = 8.0 m/s.
a) How far does he jump in the horizontal direction?
(Assume his motion is equivalent to that of a particle.)
b) What is the maximum height reached?
vxi = vi cos(θi) = 7.5 m/s
vyi = vi sin(θi) = 4.0 m/s
R = 7.94 m
b) What is the maximum height?
h = [(vyi)2/(2g)]
h = 0.72 m
Kinematic Equations
vxi = vicosθi, vyi = visinθi ,vxf = vxi
xf = vxi t, vyf = vyi – gt
yf = vyi t - (½)gt2
(vyf) 2 = (vyi)2 - 2gyf
Example: Driving Off a Cliff!!
A movie 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?
Kinematic Equations: vxi = vicosθi, vyi = visinθi ,vxf = vxi xf = vxi t
vyf = vyi – gt, yf = vyi t - (½)gt2, (vyf) 2 = (vyi)2 - 2gyf
vxf = vxi = ? vyf = -gt
xf = vxft, yf = - (½)gt2
Time to Bottom:
t = √2y/(-g) = 3.19 s
vx0 = (x/t) = 28.2 m/s
Solutions: Driving Off a Cliff!!
A movie 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?
Kinematic Equations: vxi = vicosθi, vyi = visinθi ,vxf = vxi xf = vxi t
vyf = vyi – gt, yf = vyi t - (½)gt2, (vyf) 2 = (vyi)2 - 2gyf
vxf = vxi = ? vyf = -gt
xf = vxft, yf = - (½)gt2
Time to Bottom:
t = √2y/(-g) = 3.19 s
vx0 = (x/t) = 28.2 m/s
vx = vxi = ?, vyf = -gt
xf = vxit, yf = - (½)gt2
Time to the bottom =
time when y = - 50 m
- (½)gt2 = - 50 m
t = 3.19 s
At that time xf = 90.0 m
So vxi = (xf/t) = (90/3.19)
vxi = 28.2 m/s
Example: Kicked Football
• A football is kicked at an angle θ0 = 37.0° with a velocity of
20.0 m/s, as shown. Calculate:
a. The maximum height. b. The time when it hits the ground.
c. The total distance traveled in the x direction.
d. The velocity at the top. e. The acceleration at the top.
θ0 = 37º, v0 = 20 m/s
 vx0= v0cos(θ0) = 16 m/s, vy0= v0sin(θ0) = 12 m/s
Conceptual Example
vx0 
Demonstration!!
Conceptual Example: Wrong Strategy
“Shooting the Monkey”!!
Video Clips!!
Example: A Punt!
vi = 20 m/s, θi = 37º
vxi = vicos(θi) = 16 m/s, vyi= visin(θi) = 12 m/s
Proof that the projectile path is a parabola
xf = vxi t , yf = vyi t – (½)g t2
Note: The same time t enters both equations!
 Eliminate t to get y as a function of x.
Solve the x equation for t: t = xf/vxi
Get: yf = vyi (xf/vxi) – (½)g (xf/vxi)2
Or: yf = (vyi /vxi)xf - [(½)g/(vxi)2](xf)2
This is of the form yf = Axf – B(xf)2
A parabola in the x-y plane!!
Example : The Stranded Explorers
Problem: An Alaskan rescue plane drops a package of emergency rations
to a stranded party of explorers, as shown in the picture. If the plane is
traveling horizontally at vi = 42.0 m/s at a height h = 106 m above the
ground, where does the package strike the ground relative to the point at
which it is released?
vi = 42 m/s
h
Problem
Problem Solution
Choose the origin at ground level, under where the projectile is launched, &
up to be the positive y direction. For the projectile:
v0  65.0 m s ,
0  35.0, a y   g , y0  115 m, v y 0  v0 sin  0 .
a. The time to reach the ground is found from the free fall equation, with
final height = 0. Choose positive time since the projectile was launched at t = 0.
y  y0  v y 0t  12 a y t 2
t

0  y0  v0 sin  0t  12 gt 2
 v0 sin  0  v02 sin 2  0  4   12 g  y0
2  g 
1
2

 9.964 s ,  2.3655s  9.96s
b. The horizontal range is found from the horizontal motion at
constant velocity.
v x  v0 cos  0   65.0 m s  cos 35.0  53.2 m s .
x  v x t   v0 cos  0  t   65.0 m s   cos 35.0   9.964 s   531m
c. At the instant just before the particle reaches the ground, the
horizontal component of its velocity is the constant
v x  v0 cos  0   65.0 m s  cos 35.0  53.2 m s .
The vertical component of velocity is found from:

v y  v y 0  at  v0 sin 0  gt   65.0 m s  sin 35.0  9.80 m s 2
 9.964 s 
 60.4 m s
d. The magnitude of the velocity is found from the x and y
components calculated in part c. above.
v
v x2  v 2y 
 53.2 m s    60.4 m s   80.5 m s
2
2
e. The direction of the velocity is
  tan
1
vy
vx
so the object is moving
 tan
1
60.4
53.2
 48.6
48.6 below the horizon .
f. The maximum height above the cliff top reached by the projectile
will occur when the y-velocity is 0, and is found from:
v 2y  v 2y 0  2a y  y  y0 
ymax 
v02 sin 2  0
2g
0  v02 sin 2  0  2 gymax

 65.0 m s  sin 2 35.0
2


2 9.80 m s
2

 70.9 m