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Projectile motion - REG
- 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
 Maximum height ymax = H
max height: y = uy t +
y=
at the top vy = 0
1 2
gt
2
uy  v y
t
vx = ux
x = ux t
y:
vy = uy + gt
y = uy t +
g 2
t
2
u  vy
y= y
t
2
v2y =u2y + 2gy
vertical component of the velocity is zero, vy = 0.
example: object is thrown into the air with initial velocity
uy = u sin 300 = 8 m/s
x:
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
H=
8+0
0.8 = 3.2m
2
2
 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:
x = ux t
u = 16 m/s, 300 . Find how far did it land..
t  8 - 5t  = 0
1. t = 0 corresponds to x = 0
ux = u cos
300
and
= 6.9 m/s
that motion is totally symmetrical
v y = uy + gt = 8 - 10  1.6 = - 8 m/s = -uy and vx = ux
2. 8 – 5 t = 0
t = 1.6 s
R = (6.9) (1.6) = 11.4 m/s
2
 How does the range depend on the angle
of initial velocity:
maximum range for 0 = 450.
any 0 will have the same range as 900 – 0.
 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.
 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
y = 4.9x 0.5 – 4.9 x 0.52 = 1.2m
How much time he spent in the air? t = 2 x 0.5 s = 1 s
How far did he land?
 t = 0.5 s
x = ux t = 8.5x1 = 8.5 m
(No, no he didn’t)