This problem assumes familiarity with the earlier example problem, Throwing a Baseball 1 (The Basics).
Again suppose you are throwing a baseball, just as we assumed in Parts A and B of the earlier example. You release the ball with a perfectly horizontal velocity at a height of 1.5 m above the ground. The ball travels 5.0 m horizontally from the instant it leaves your hand until the instant it first contacts the ground. How fast is the ball moving just before it impacts the ground?
Solution
System, Interactions and Models: As in Parts A and B of the earlier example
Approach:
The givens are the same as in Part B of the earlier example. Defining the same coordinate system as was used in that Part, we have:
Further, we can follow the same steps as we did in Part B from before to find the x-velocity.
This time, however, we are not done yet. The x-velocity is only half of the story. When the ball is first released, its
velocity is entirely in the x direction, but during its flight it acquires a y-component.
When asked how fast something is traveling, we generally make no distinction between x- and y-components. Instead, we want to know the size of the total velocity. To find the total velocity, we must use the Pythagorean theorem:
There are many ways to find the y-component of the velocity. One possibility is to use the equation:
which, after substituting zeros gives:
In this case, we must choose the negative sign. This is true because the ball is clearly moving downward (as opposed to upward) just before striking the ground. Thus, the vertical component of the ball is negative in our coordinate system.
We therefore find that:
Note that we did NOT choose between plus or minus this time. The Pythagorean theorem does not have one. The reason for this is that the Pythagorean theorem gives lengths (when dealing with its orginal application), or in the case of vectors like velocity or acceleration, magnitudes. Magnitudes can never be negative.
This leads us to an important aside. Whenever the Pythagorean theorem is being used, we are clearly working with vectors in two or more dimensions. Once we have moved beyond one dimension, it is impossible to unambiguously report the direction of a vector by assigning a plus or minus sign to the magnitude. The baseball of this problem is a good example. It is clearly moving to the right (positive in our coordinate system) and down (negative in our coordinate system).
To illustrate the appropriate way to report direction, let us now find the velocity of the ball just before contacting the ground.
We have the vector triangle:
Note that we have removed the negative sign from the y-component. This is correct, because the arrow is used in the drawing to denote the direction. An arrow and a negative sign can result in confusion, in just the same way as reporting a velocity of " – 2 m/s west" can result in confusion.
We can now find the angle shown in the figure by using trigonometry. One possibility is to use:
Can you get the same result for θ using the arccosine or arcsine functions?
Reporting the vector is a little tricky here, since we are not told which direction the ball is thrown (north? south? east? west?). The best we can do is to say the final velocity is 10.5 m/s at 31° below the horizontal.