Projectile Motion Notes
Projectile Motion
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A vector can be broken down into
component vectors by creating a right
triangle. This is done to evaluate
motion in two dimensions.
Projectile Motion
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In order to determine the magnitude and
direction of the component vectors,
Trigonometry and the Pythagorean
Theorem must be used.
Projectile Motion
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When an object is thrown or launched
into the air, it undergoes projectile
motion. The pathway of any projectile
is a parabola (excluding air resistance).
Projectile Motion
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After breaking the initial velocity into
x and y components separately, the
pathway of the object can be tracked
by using a series of equations:
Projectile Motion Equations
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Variables:
Y = distance in meters vertically
Vy = y component of velocity
Voy = y component of initial velocity
Vox = x component of initial velocity
g = acceleration of gravity
t = time during flight
Projectile Motion Equations
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Vertical:
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Velocity:
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Distance:
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Time for maximum height:
vy = vo(y) – g t
Y = vo(y) t – ½ g t2
t = vo(y) / g
Projectile Motion Equations
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Horizontal:
Velocity: Vx remains constant;
therefore vo(x) = vf (x)
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Distance: X = vo(x)
t
Projectile Motion
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In projectile problems, vertical and
horizontal motions are completely
independent of one another.
For example: An object will accelerate
downward at the exact same rate regardless
of whether it is dropped, thrown forward
slowly, or thrown forward very quickly.
Likewise: An object will continue to move
forward the same distance every second
whether it is rolling along a flat surface or
moving vertically.
Projectile Motion
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Time is the only variable which will be the
same for the two motions. The time it takes
for an object to hit the ground is the same
amount of time it has to move forward.
Projectile Motion Equations
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One method for solving for horizontally
projected objects:
1) Write down all variables, and identify
them as horizontal or vertical.
2) Solve one dimension (x or y, depending on
what is given in the problem) for time.
3) Use that time in the other dimension to
find unknown quantity
Practice Problems
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Turn to page 129, work through problem on
right side of page
Asking for vertical distance (y) and horizontal
distance (x)
Initial vertical velocity = 0 m/s, Initial
horizontal velocity = 20 m/s, t = 2 seconds,
g = -9.81 m/s2
Solve for y, solve for x