### Video Transcript

A tennis ball of mass 150 grams is
thrown at an angle of 40 degrees above the horizontal and a speed of 18 meters per
second. What is the momentum of the ball
after it has been in flight for 0.36 seconds?

We can call the ballโs momentum
after itโs been in flight for this amount of time ๐ and begin on our solution by
drawing a diagram of this situation. Our sketch represents that weโve
thrown this ball at an angle, weโve called ๐, of 40 degrees above the horizontal at
an initial speed, weโve called ๐ฃ sub ๐, of 18 meters per second. After a time in flight, weโve
called ฮ๐ก, of 0.36 seconds, we want to know the momentum of the ball.

If we let upward motion be motion
in the positive ๐ฆ-direction and motion to the right be in the positive ๐ฅ and if we
recall that momentum ๐ is equal to an objectโs mass times its velocity, then we can
write that momentum ๐ is equal to our ballโs mass ๐ multiplied by the square root
of ๐ฃ sub ๐ฅ squared plus ๐ฃ sub ๐ฆ squared. Here, ๐ฃ sub ๐ฅ is the ballโs speed
in the ๐ฅ-direction at a time value of 0.36 seconds. And likewise, ๐ฃ sub ๐ฆ is the
ballโs vertical speed at that same time.

Since weโve been told the ballโs
mass ๐, that means if we can solve for ๐ฃ sub ๐ฅ and ๐ฃ sub ๐ฆ, weโll be able to
solve for momentum ๐. Starting with ๐ฃ sub ๐ฅ, when we
look at our diagram, we see that we can represent the speed of the ball in the
๐ฅ-direction in terms of ๐ฃ sub ๐ and ๐. ๐ฃ sub ๐ฅ which, recall, is the
ballโs ๐ฅ-velocity component at time ๐ก equals 0.36 seconds is equal to the ballโs
initial launch speed ๐ฃ sub ๐ times the cosine of the angle ๐.

Plugging in for these values, when
we calculate ๐ฃ sub ๐ฅ, we find itโs approximately 13.79 meters per second. This is the ballโs speed in the
๐ฅ-direction and is constant throughout the ballโs flight. Now that we know ๐ฃ sub ๐ฅ, we move
on to solving for ๐ฃ sub ๐ฆ, the ballโs vertical speed component at time ๐ก equal
0.36 seconds.

Unlike the ballโs ๐ฅ-component
speed, its ๐ฆ-component speed is not constant in time, but constantly changes under
the influence of gravity. As a projectile, the ballโs motion
can be described by the kinematic equations. In particular, we recall the
relationship final speed is equal to initial speed plus acceleration times time. When we write this equation in
terms of our variables, we can write that ๐ฃ sub ๐ฆ is equal to the initial speed of
the ball in the ๐ฆ-direction minus the acceleration due to gravity ๐ times ฮ๐ก.

Weโll treat gravitational
acceleration ๐ as exactly 9.8 meters per second squared. Looking again at our diagram, we
see that we can write the expression for ๐ฃ sub ๐ฆ๐, the initial speed of the ball
in the vertical direction, as the product of ๐ฃ sub ๐, the ballโs initial speed,
times the sin of the angle ๐. Since weโre given ๐ฃ sub ๐, ๐,
and ฮ๐ก, and ๐ is a constant, weโre ready to plug in and solve for ๐ฃ sub ๐ฆ.

When we do and enter this
expression on our calculator, we find that the ๐ฃ sub ๐ฆ is approximately 8.042
meters per second. Now that we know ๐ฃ sub ๐ฅ and ๐ฃ
sub ๐ฆ, weโre ready to plug in and solve for momentum ๐. When we do plug in, weโre careful
to convert our mass into units of kilograms. And we find that ๐ is equal to 2.4
kilograms meters per second. Thatโs the overall momentum of this
ball 0.36 seconds after its been launched.