Projectile Motion - Angled Launches
10 m/s
5 m/s
Fill in the component and resultant velocities of the
thrown projectile as it ascends and descends.
11.2 m/s
Assume that g = - 10 m/s2
5 m/s
20 m/s
Determine the maximum height reached by the
projectile
11.2
20.6 m/s
vvi = 30 m/s vvtop = 0 m/s
g = -10 m/s2 H = ?
vvf2 = vvi2 + 2gH
H = (vvtop2 - vvi2) / 2g
H = (0m/s2 - 30m/s2) / 2(-10m/s2) = 45m
20.6
Determine the time in the air and the range
30.4 m/s
vvf = vvi + gt
But….vvf = - vvi
So ….-vvi = vvi + gtAir
-2vvi = gtAIr So.. tAir = -2vvi / g = -2(30m/s)/ -10m/s2 = 6s
dH = vH t
30.4
R = vH tAir = (5m/s) (6s)
= 30m
Projectile Motion - Angled Launches
t = 2s
t = 1s
5m
20m
Projectile Motion - Angled Launches
Projectile Motion - Angled Launches
If there wasn’t any gravity the banana
would follow a straight line trajectory and
of course reach the monkey which would
also not fall below its inertial position
If there is gravity the banana falls below
its inertial position by 1/2gt2. If the
monkey lets go at the same time the
banana is fired, he too will fall below his
rest position by 1/2gt2.
Projectile Motion - Angled Launches
If the keeper fires the banana at a slower
speed it will still fall below its inertial position
by 1/2gt2 but the horizontal component of
its speed will be smaller so the banana will
take longer to reach the monkey. If too slow
the monkey will hit the ground before the
banana can get across.
Projectile Motion - Angled Launches
Conceptual Example
I Shot a Bullet into the Air...
Suppose you are driving a convertible with the top down.
The car is moving to the right at constant velocity. You point
a rifle straight up into the air and fire it. In the absence of air
resistance, where would the bullet land – behind you, ahead
of you, or in the barrel of the rifle?
Bullet retains the horizontal
velocity of the car so relative to a
person in the car it will appear to
go straight up and down.
Example The Height of a Kickoff
A placekicker kicks a football at and angle of 40.0 degrees and
the initial speed of the ball is 22 m/s. Ignoring air resistance,
determine the maximum height that the ball attains.
vo

voy
vox
voy  vo sin    22m s  sin 40  14 m s
vox  vo cos   22m s  cos40  17 m s
y
ay
vy
voy
?
- 9.8 m/s2
0
14 m/s
t
y
ay
vy
voy
?
-9.80 m/s2
0
14 m/s
v  v  2ay y
2
y
2
oy
y
y
0  14 m s 

2 9.8m s
2
2

 10 m
t
v v
2
y
2a y
2
oy
The Time of Flight of a Kickoff
What is the time of flight between kickoff and landing?
y
ay
0
-9.80 m/s2
vy
voy
t
14 m/s
?
y
ay
vy
0
-9.80 m/s2
voy
t
14 m/s
?
y  voyt  ayt
1
2

2

0  14m st   9.80m s t
1
2

2
2

0  214m s   9.80m s t
t  2.9 s
2
The Range of a Kickoff
Calculate the range R of the projectile.
x  voxt  17m s  2.9 s   49 m
Conceptual Example
Two Ways to Throw a Stone
From the top of a cliff, a person throws two stones. The stones
have identical initial speeds, but stone 1 is thrown downward
at some angle above the horizontal and stone 2 is thrown at
the same angle below the horizontal. Neglecting air resistance,
which stone, if either, strikes the water with greater velocity?
Both stones strike
the water with the
same velocity
Shoot the Monkey - Problem
A projectile is shot at the center of a target 2m away and 4m high. The target
drops at the same time the projectile leaves the gun and is hit after it falls
1m. Determine the initial velocity of the projectile (speed and angle ).
1m
H =4m
dv
vi

dH = 2m
Shoot the Monkey - Solution
Horizontal
dH = 2m
Vertical
dv = 3m
dH = vH t = (vi cos) t
g = -9.8 m/s2
dv = vvi t + 1/2 g t2
dv = (vi sin) t + 1/2 g t2
t = dH
(vi cos)
dv = (vi sin) dH
(vi cos)
 = tan-1 (H /dH)
= tan-1 (4m /2m)
+ 1/2 g ( dH )2
vi cos
= 63.430
Shoot the Monkey - Solution
dv = (sin) dH
(cos)
+ 1/2 g ( dH )2
vi cos
3m = (tan 63.430) (2m)
3m = 3.999m
+
+ 1/2 (- 9.8 m/s2) (
- 97.966 m3/s2
v i2
- 0.999 m = -97.966 m3/s2
v i2
vi =  (- 97.966 m3/s2 )
-0.999m
= 9.9 m/s
2m
)2
vi cos 63.430
Shoot the Monkey - Check
The projectile “falls” below its path (dashed) due to gravity. The freefall distance (dFF)
is therefore 1m.
Horizontal
dH = 2m
Vertical
dFF = 1m
dH = vH t
vH = dH
t
vH = 2m = 4.43 m/s
0.45s
dFF = 1/2 g t2
t =  2 (dFF) / g
t =  2(1m) / 9.8m/s2
= 0.45 s
Shoot the Monkey - Check
vi
vvi
 = tan-1 (H / dH)
= tan-1 (4m / 2m)
 = 63.430

Cos  = vH / vi
vH
vi = vH / Cos 
vi = 9.9 m/s
= 4.43m/s / Cos (63.430)