Firing a cannonball A 5.0 kg cannonball is fired straight up at 35 m/s. What is its speed after rising 45 m? MODEL Let the system consist of only the cannonball, which we model as a particle. Assume that air resistance is negligibl

Question

Firing a cannonball

A 5.0 kg cannonball is fired straight up at 35 m/s. What is its speed after rising 45 m?

MODEL Let the system consist of only the cannonball, which we model as a particle. Assume that air resistance is negligibl

VISUALIZE FIGURE 9.3 is a before-and-after pictorial representation.
Because before-and-after representations are usually simpler than the pictorial representation used in dynamics problems, you can include known information right on the diagram instead of making a Known table.

Accepted Answer

It isn’t necessary to use work and energy to solve this problem. You could solve it as a free-fall problem. Or you might have previously learned to solve problems like this using potential energy, a topic we’ll take up in the next chapter. But using work and energy emphasizes how these two key ideas are related, and it gives us a simple example of the problem-solving process before we get to more complex problems. The energy principle is ∆K = W, where work is done by the force of gravity. The cannonball is rising, so its displacement ∆y is positive. But the force vector points down, with component [latex]F_{y} = -mg[/latex]. Thus gravity does work
[latex]W = F_{y} ∆y = -mg ∆y = -(5.0 kg)(9.80 m/s^{2})(45 m) = -2210 J [/latex]
as the cannonball rises 45 m. A negative work means that the system is losing energy, which is what we expect as the cannonball slows

The cannonball’s change of kinetic energy is [latex]∆K = K_{f} - K_{i}[/latex].
The initial kinetic energy is
[latex]K_{i} = \frac{1}{2} m v_{i}^{2} = \frac{1}{2} (5.0 kg)(35 m/s)^{2} = 3060 J[/latex]
Using the energy principle, we find the final kinetic energy to be [latex]K_{f} = K_{i} + W = 3060 J - 2210 J = 850 J[/latex] . Then

[latex]v_{f} = \sqrt{\frac{2K_{f}}{m}} = \sqrt{\frac{2(850 J)}{5.0 kg}} = 18 m/s [/latex]
ASSESS [latex]35 m/s \approx 70 mph[/latex]. A cannonball fired upward at that speed is going to go fairly high. To have lost half its speed at a height of
[latex]45 m \approx 150 ft[/latex] seems reasonable.

Question 9.1: Firing a cannonball A 5.0 kg cannonball is fired straight up...

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