2 and 3 DIMENSIONAL MOTION

 dv
a
dt

 dr
v
dt
Constant acceleration
 2
  
at
r  r0  v 0 t 
2
 

v  v 0  at


 v0  v
r 
t
2
 
2ar  v 2  v 02
t  t  t0
Projectile Motion
The trajectory of an object projected with an initial

velocity v0 at the angle  0 above the horizontal
with negligible air resistance.
Example: A projectile is fired from a cannon at a 30-degree angle
with the ground and an initial velocity of 100 m/s. Assuming no air
resistance and g = 10 m/s2, calculate the time it will spend in the air.
a. 2.5 s
y
b. 5.0 s
c. 10 s
d. 20 s
e. 40 s
v0 = 100 m/s
θ = 30
yfinal = 0
1
y  y 0  v 0yt  ayt 2
2
0  0  v 0 sin t 
1 2
gt
2
x
t  0 (start!)

2v 0
2 100 m/s 

t

sin


sin30  10 s

2
g
10 m/s

Example
y

v0
g

v0  v0 x , v0 y 
v0  v02x  v02y
v0 x  v0 cos 
x

x  v0 x t
t
gt 2
y  v0 y t 
2
v x  v0 x
tan  
v0 y  v0 sin 
x
v0 x
Tragectory :
v0 y
g
y
x  2 x2
v0 x
2v 0 x
v y  v0 y  gt
x  xmax  y  0  t0  0, and t 
xmax 
2v0 x v0 y
g
g

2v02 sin  cos  v02 sin 2


g
g
v02
  45  max xmax  
g

2v0 y
v0 y
v0 x
Shoot the monkey (tranquilizer gun)
A zookeeper shoots a tranquilizer dart to a monkey that hangs from a
tree. He aims at the monkey and shoots a dart with an initial speed v0.
The monkey, startled by the gun, lets go immediately.
Will the dart hit the monkey?
A. Only if v0 is large enough.
B. Yes, regardless of the magnitude of vo.
C. No, it misses the monkey.
If there is no gravity, the
dart hits the monkey…
r m  r m0
If there is gravity, the dart
also hits the monkey!
r m  r m0 
r b  r b0  vb0 t
1 2
gt
2
1 2
r b  r b 0  v b 0 t  gt
2
Continued
If there is no gravity, the dart hits the monkey…
r m  r m0
r b  r b0  vb0 t
Continued
If there is gravity, the dart also hits the monkey!
1 2
r m  r m 0  gt
2
1 2
r b  r b 0  v b 0 t  gt
2
Note, that it takes the same amount of time to hit the monkey
as in the no gravity case!
Continued
This might be easier to think about…
x  v0t
For the bullet:
1 2
y  gt
2
x  x0
For the monkey:
1 2
y  gt
2
Example: A ball is hit from a platform, 1.0 m above the ground, with an
initial velocity of 36.5 m/s and at an angle of 30° above horizontal. A 3 m
high fence is 113 m from the base of the platform. Neglect air resistance.
a)
b)
c)
How long after the hit does the ball reach the fence?
Does the ball go over the fence?
What is the speed of the ball as it hits the ground?
x0  0
y 0  1m
v0  36.5m / s
y
  30 
θ
1m
x  x0  v0 x t
3m
Fence

gt 2
y  y 0  v0 y t 

2
t
x  113m
h  3m
t  ? y  ? v y 0  ?
x
x  x0
x  x0
113m


 3.57 s

v0 x
v0 cos  36.5m / s  cos 30


y  1.0m  36.5m / s sin 30   12 9.8m / s 2 3.57 s   3.5m
v 2  v02  2 g ( y  y 0 )  v y 0  v02  2 gy 0 
2
36.5m/s 2  29.8m / s 2 1m
 36.8m / s
Example (continued): An alternative (and waaaaay longer) approach for part (c):
y  0:
0  y 0  v 0 yt 
1
gt 2
2
1
0  1  (36.5 m/s) sin30t  (9.81 m/s2 )t 2
2

t  3.77 s
v x  v 0x  v 0 cos   (36.5 m/s)cos30  31.6 m/s
v y  v 0 sin   gt  (36.5 m/s) sin30  (9.81 m/s2 )(3.77 s)  18.8 m/s
v  vx2 vy2  36.8 m/s
Example1: A battleship simultaneously fires two shells at enemy ships. If the
shells follow the parabolic trajectories shown, which ship gets hit first?
tA
Shell A spends 2tA in the air, where tA is the time
it takes for vy to become zero: 0 = v0Ay — gtA.
tA
1. A
Shell A goes higher
v0Ay > v0By
tA > tB
The vertical part of the motion dictates the time
a projectile spends in the air.
2. both at the same time
3. B
4. need more information
Example2: Two snowballs have been thrown with initial speed v0, at angles
θ1 = 67.5° and θ2 = 22.5°. Find the ratio of their ranges, R1/R2 and the ratio of
their time in flight t1/t2.
From an example in
the previous lecture:
2v0 sin 
t1 sin 1 sin 67.5 
t

 

 2.4

g
g
t 2 sin  2 sin 225
2v 0 y
x max
v02 sin 2
R1 sin 21 sin 135 




1

g
R2 sin 2 2
sin 45