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Projectile motion - HON
- motion of objects (thrown or projected into the air with an initial velocity
gravitational force if we can neglect air resistance.
u ) upon which the only force is
ux = u cos 0
uy = u sin 0
The approximation of neglecting air resistance is not generally justified, especially at high velocities. In addition a
spin of a projectile – baseball – can give rise to some effects associated with aerodynamic forces.
- We consider only objects near the Earth’s surface, so g doesn’t change too much in direction and magnitude.
With these assumptions projectile trajectory (path) is parabolic..
DEFINITION: Projectile motion is combined motion of the
two independent motions simultaneously
– one in horizontal (x) direction with constant velocity
vx = ux (no force in x direction  no acceleration
 no change in velocity in that direction)
– one in vertical (y) direction with constant
acceleration g downwards
HORIZONTAL MOTION
ux = u cos 0
vx = ux
x = ux t
VERTICAL MOTION
uy = u sin 0
vy = uy + gt
v2y =u2y + 2gy
y = uy t +
g 2
t
2
y=
uy  v y
2
t
With a little help of kinematic equations for 2-D motion ,
we know everything about that motion is we know initial velocity u, 0.
x = ux t
 t = x/ux
y =  tan 0  x +
into y = uy t +
g
x2
2u cos2 0
2
g 2
t
2
 y=
uy
ux
x+
g 2 u sin0
g
x 
x + 2 2 x2
2u2x
u cos0
2u cos 0
That trajectory is parabola in Galileo’s days was forefront of physics research.
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 Maximum height ymax = H
vertical component of the velocity is zero at the top, vy = 0.
example: object is thrown into the air with initial velocity
uy = u sin 300 = 8 m/s
at the top vy = 0
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max height: y = uy t + gt 2
2
y=
uy  v y
2
t =
u = 16 m/s, 300. Find maximum height.
uy + gt = 0
8 – 10t = 0
t = 0.8 s
H = 8 x 0.8 – ½ x 10 x 0.82 = 3.2 m or
8+0
2
0.8 = 3.2 𝑚
H = 3.2 m
 Horizontal Range xmax = R For a projectile beginning and ending at the same height, the time it takes to rise to its highest
point equals the time it takes to fall back to the original height. At that point y = 0.
example: object is thrown into the air with initial velocity
y = uy t +
1 2
gt = 0
2
two solutions:
for final point y = 0
t  8 - 5t  = 0
1. t = 0 corresponds to x = 0
x = ux t
ux = u cos
300
u = 16 m/s, 300 . Find how far did it land..
= 6.9 m/s
and
2. 8 – 5 t = 0
x = (6.9)(1.6) = 11.4 m
t = 1.6 s
R = 11.4 m
or
from previous example time for reaching the top is 0.8 s, so the whole flight is 1.6 s: x = ux t = 11.4 m
 How does the range depend on the angle
of initial velocity:
Range R (x = R, y = 0 )
0 = uyt + (g/2) t2
t = 2uy/g
R = ux t = 2 ux uy/g = 2 u2 sin  cos 
maximum range for  = 450.
sin 2 = 2 sin  cos  , therefore any  will have the same range as 900 – .
 Projectile motion with air resistance
 Objects dropped from a moving vehicle have the same horizontal velocity as the moving vehicle. In general, object dropped from
anything that is moving with certain velocity will have initial velocity in the air equal to the velocity of the moving object. For
example f you walk and drop a pen, that pen will have horizontal velocity equal to your horizontal velocity. Only now in the air it
will have vertical velocity too.
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 Example: A projectile is launched with an initial velocity with: vertical component of 40 m/s (uy = 40 m/s)
and horizontal component of 30 m/s (ux = 30 m/s)
Find: a) time need for max height b) velocity at max height c) max height d) horizontal max distance if it lands at the same height
as it was launch e) total time of flight
a) after 4s vertical velocity is 0, because it decreases 10 m/s every second.
or math: at the top vy = 0, so uy + gt = 0 ; 40 – 10t = 0 or t = 4s
b) At max height velocity is completely horizontal: v = ux = 30 m/s;
40+0
1 2
y = 40x4 – ½ x10x42 = 80 m or y =
4 = 80m
gt
2
2
d) In that time the object moved (30m/s)(8s) = 240m in horizontal direction.
c) max height: y = uy t +
e) Total time of flight: 4s x 2 = 8 s.
 Example: Harry accidentally falls out of a helicopter that is traveling horizontally at 60 m/s.
He plunges into the water below 3.00 seconds later. Assuming no air resistance, what is the horizontal
distance he travels while falling? How high was helicopter?
ux = 60 m/s
x = ux t = 180 m
uy = 0
y = uy t +
1 2
gt
2
y = 0 + 5 x 9 = 45 m
 Example: Jack be nimble, Jack be quick, Jack jumped over the candlestick with a velocity of 9.8 m/s at
an angle of 30.0° to the horizontal. Did Jack burn his feet on the 1 m high candle?
ux = 9.8 cos 30 = 8.5 m/s
ux = 9.8 sin 30 = 4.9 m/s
at the top: vy = 0  uy + gt = 0  4.9 – 9.8 t = 0
y = uy t +
1 2
gt
2
 t = 0.5 s
y = 4.9x 0.5 – 4.9 x 0.52 = 1.2m
(No, no he didn’t)
How much time he spent in the air? t = 2 x 0.5 s = 1 s
What was his velocity 0.7 s after jump?
v = v2x +v2y
θ = arc tan
How far did he land?
vx = ux = 8.5 m/s
vy = uy + gt = 4.9 – 9.8x0.7 = – 1.96 m/s
v = 8.7 m/s
1.96
= 130
8.5
x = ux t = 8.5x1 = 8.5 m
v = 8.7m/s, -130
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PROJECTILE MOTION HW2 - HONOURS
1. Harry accidentally falls out of a helicopter that is traveling at 100 m/s. He plunges into a swimming pool 2 seconds later.
Assuming no air resistance, what was the horizontal distance between Harry and the swimming pool when he fell from the
helicopter?
2. Harry and Angela look from their balcony to a swimming pool below that is 15 m from the bottom of their building. They
estimate the balcony is 45 m high and wonder how fast they would have to jump horizontally to succeed in reaching the pool.
What is your answer?
3. A boy on the tower in the figure below throws a ball a distance of 60 m, as shown. At what speed, in m/s, is the ball thrown?
4. A projectile is fired over level ground with an initial velocity that has a vertical component of 20 m/s and a horizontal component
of 30 m/s. Using g = 10 m/s2, the distance from launching to landing points is:
5. The Essex county sheriff is trying to determine the speed of a car that slid off a small bridge on a snowy New England night and
landed in a snow pile 4.00 m below the level of the road. The tire tracks in the snow show that the car landed 12.0 m measured
horizontally from the bridge. How fast was the car going when it left the road?
6. Emanuel Zacchini, the famous human cannonball of the Ringling Bros. and Barnum & Bailey Circus, was fired out of a cannon
with a speed of 24.0 m/s at an angle of 40.0° to the horizontal. If he landed in a net 56.6 m away at the same height from which he
was fired, how long was Zacchini in the air?
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7. A projectile is launched at an angle into the air.
a. Neglecting air resistance, what is its vertical acceleration?
b. Its horizontal acceleration?
c. At what point in its path does a projectile have minimum speed?
8. Why is it important that such a satellite be above Earth’s atmosphere?
9. What force acts on a satellite that is above Earth’s atmosphere?
10.At her wedding Jennifer lines up all the single females in a straight line away from her in preparation for the tossing of the bridal
bouquet. She stands Kelly at 1.0 m, Kendra at 1.5 m, Mary at 2.0 m, Kristen at 2.5 m, and Lauren at 3.0 m. Jennifer turns around and
tosses the bouquet behind her with a speed of 3.9 m/s at an angle of 50.00 to the horizontal, and it is caught at the same height
0.60 s later. a) Who catches the bridal bouquet? b) Who might have caught it if she had thrown it more slowly?
11. Mubarak jumps and shoots a field goal from the far end of the court into the basket at the other end, a distance of 27.6 m. The ball
is given an initial velocity of 17.1 m/s at an angle of 40.0° to the horizontal from a height of 2.00 m above the ground. What is its
velocity as it hits the basket 3.00 m off the ground?
12. Drew claims that he can throw a dart at a dartboard from a distance of 2.0 m and hit the 5.0-cm-wide bulls-eye if he throws the dart
horizontally with a speed of 15 m/s. He starts the throw at the same height as the top of the bulls-eye. See if Drew is able to hit the
bulls-eye by calculating how far his shot falls from the bulls-eye’s lower edge.
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13. Caitlin is playing tennis against a wall. She hits the tennis ball from a height of 0.5 m above the ground with a velocity of 20.0 m/s at
an angle of 15.0° to the horizontal toward the wall that is 6.00 m away. a) How far off the ground is the ball when it hits the wall? b)
Is the ball still traveling up or is it on its way down when it hits the wall?
14.A stone is thrown at an angle of 30.00 above the horizontal from the top edge of a cliff with an initial speed of 12.0 m/s. A stopwatch
measures the stone’s trajectory time from the top of the cliff to the bottom at 5.6 s. What is the height of the cliff? (Disregard air
resistance.) What is the speed of the stone at the bottom?
15.A stone is thrown horizontally from the top of a 20.0-m high hill. It strikes the ground at an angle of 450. With what speed was it
thrown?
16. Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly for lunch. If the lily pads are spaced 2.4 m apart, and
Ferdinand jumps with a speed of 5.0 m/s, taking 0.60 s to go from lily pad to lily pad, at what angle must Ferdinand make each of
his jumps?
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SOLUTIONS:
1. x = vx t = 200 m
2. vx = x/t = 5 m/s (because t = 3s for a 45-m vertical fall)
3. 30 m/s (it takes 2 s to fall 20 m)
4. 120 m
5.
t
y = (g/2) t2
2y
g
t = 0.894 s
ux = x/t
ux = 13.4 m/s
6. Because Zacchini was in the air for the same amount of time vertically that he was horizontally, you can find his horizontal time
and this will be the answer. First, you need the horizontal velocity component.
ux = u cos  = (24m/s) cos 400 = 18.4 m/s
Now you have the horizontal velocity component and the horizontal displacement, so you can find the time.
Given: ux = 18.4 m/s
x = 56.6 m so
x = ux t t = 3.08 s
7. a. Its vertical acceleration is g because the force of gravity is downward.
b. Its horizontal acceleration is zero because no horizontal force acts on it.
c. The minimum speed of a projectile occurs at the top of its path. if it is launched vertically, its speed at the top is zero. If it is
projected at an angle, the vertical component of velocity is still zero at the top, leaving only the horizontal component. So the
speed at the top is equal to the horizontal component of the projectile's velocity at any point.
8. To be above the effect of drag, which would reduce its speed.
9. only gravity
10. ux = (3.9 m/s) cos 50.00
ux = 2.5 m/s
x = ux t
x = 1.5 m a. Kendra b. Kelly
11. uy = u sin  = 11.0 m/s
vy 2 = uy 2 + 2gy
12. t = x/ux = (2.0 m)/(15 m/s) = 0.13 s
13. a. ux = u cos  = 19.3 m/s
y = uy t + ½ g t2 = 1.13 m
+
uy = u sin  = 6.00 m/s
uy = u sin  = 5.18 m/s
t = x/ux = 0.311 s
ytotal = 1.13 + 0.50 m = 1.63 m
ux = u cos  = 10.4 m/s
15.
+
+
𝑣 = √𝑣𝑥2 + 𝑣𝑦2
uy – (10m/s2)t = - ux
16. ux = x/t = (2.4 m)(0.60 s) = 4.0 m/s
cos  = ux/u = (4.0 m/s)/(5.0 m/s) = 0.80
 = 370 to the horizontal
+
It is still going up (positive velocity)
vy = uy + gt = - 50. m/s
y = uy t + ½ g t2 = ½ g t2
vy = - ux
𝑣 = √𝑣𝑥2 + 𝑣𝑦2 = 16.4m/s
y = ½ g t2 = 0.089 m = 8.9 cm
b. vy = uy + g t = 5.18 m/s – (10 m/s2)(0.311 s) = 2.07 m/s
14.
ux = u cos  = 13.0 m/s
vy = ± 10.0 m/s
y = uy t + ½ g t2 = -120 m
v = 50.4 m/s
t
2y
 2( 20.0 m )(10.0 m / s 2 ) = 2.00 s
g
ux = 20m/s = u