Download Projectile Motion Practice

03a 1401 Projectile Motion Practice
A solid steel ball rolls in a straight line along a level
surface with constant speed. The ball then rolls off the
level surface moving through the air with negligible airresistance, subsequently striking the ground which is some
distance below the level surface.
Draw a motion diagram for the ball while on the level surface.
Use the definition of acceleration to determine the
acceleration of the ball when it is on the level surface and
behaving as described above.
a
v 0

0
t t
Use Newton’s Second Law and your previous answer to
determine the size of the net force acting on the ball on the
level surface.
F  ma  m  0  0
Sketch the trajectory of the ball all the way to the floor.
When is the ball going fastest? Slowest?
slowest at highest point, fastest at bottom right before hitting
Does the speed of the ball change continuously
while in flight? Explain.
Yes, gravity is continually changing its speed
1
Choose any two arrows from your motion diagram and find
the change in velocity vector for the two velocity arrows you
chose.
v
v
v
Use the definition of acceleration to determine the direction of
the average acceleration for the two vector velocities you
chose.

 v
a
t
therefore


a // v
Use Newton’s Second Law and your previous answer to
determine the direction of the net force acting on the ball
while in flight.


F  ma
Since mass “m” is positive, force is parallel
to acceleration, which is downward
Given the following data: The solid steel ball rolls in a
straight line along the level surface with a constant speed
of 6.0m/s. The level surface is 3.0m above ground level.
What is the ball’s launch speed? Launch angle?
vo  6.0m / s
 o  0
While the ball is in flight, what is its acceleration?
Acceleration = 9.8m/s/s downward
2
Calculate the time the ball is in flight.
y  voyt  12 a y t 2
 3.0  (6 sin 0)t  12 (9.8)t 2
 3.0  0  4.9t 2
t  0.782 s
Calculate the vertical component of the ball’s velocity at the
instant before the ball strikes the floor.
v y  voy  a y t
 0  (9.8m / s / s)(0.782s )
 7.67m / s
What is the horizontal component of the ball’s velocity at the
instant before the ball strikes the floor.
vx  vox  a x t
 6.0m / s  (0m / s / s)(0.782s)
 6.0m / s
Use the Pythagorean theorem to calculate the speed of the ball
at the instant before it strikes the floor.
v  v x2  v y2
 (6.0m / s ) 2  (7.67m / s ) 2
 9.73m / s
Calculate the “range” of the flight, i.e. the horizontal
displacement from launch to first-strike with the floor.
x  v0 xt  12 (0)t 2
 (6.0m / s cos 0)(0.782s)
 4.69m
3