acceleration caused by gravity.
Because the vertical acceleration is constant, the ball takes as much time to reach its
peak of motion as it takes to fall back to the other hand. This also means it has covered
half of the horizontal trip when it is at its peak. The path traced by the ball is a parabola,
a shape symmetrical around its midpoint.
We said the juggler chooses the y velocity so that the ball takes the right amount of time
to rise and then fall back to the other hand. What, exactly, is the right amount of time? It
is the time needed for the ball to move the horizontal distance between the juggler’s
hands.
If a juggler’s hands are 0.5 meters apart, and the horizontal velocity of the ball is
0.5 m/s, it will take one second for the ball to move that horizontal distance. (Remember
that the horizontal velocity is constant.) The juggler must throw the ball with enough
vertical velocity to keep it in the air for one second.
Projectile motion: x velocity
x velocity constant
4.8 - Sample problem: calculating initial velocity in projectile motion
Variables
What is the strategy?
- Use a linear motion equation to determine the initial vertical velocity that will result in a final vertical velocity of 0 m/s after 0.60 s (half the
total time of 1.20 s). - Use the definition of velocity to calculate the horizontal velocity, since the displacement and the elapsed time are both provided.
Physics principles and equations
We rely on two concepts. First, the motion of a projectile is symmetrical, so half the time elapses on the way up and the other half on the way
down. Second, the vertical velocity of a projectile is zero at its peak.
We also use the two equations listed below.
vyf = vyi + ayt
A juggler throws each ball so it hangs
in the air for 1.20 seconds before
landing in the other hand, 0.750
meters away.
What are the initial vertical and
horizontal velocity components?
elapsed time t = 1.20 s
horizontal displacement ǻx = 0.750 m
initial vertical velocity vy
horizontal velocity vx
acceleration due to gravity ay = í9.80 m/s^2