Work and Energy Legacy Problem #26 Guided Solution

A good problem solver reads the problem carefully, developing a mental picture of what's going on, identifies the known information and the unknown information, and then plots out a strategy using physics principles and ideas to get from the known to the unknown quantity. In this problem, we read of a baseball which leaves the bat of Dave Kingman at a speed of 92.7 meters per second. It flies through the air and clears the outfield fence, eventually hitting a house 530 feet from home plate. What we're asked to determine is the speed of the ball when it clears the stadium wall at a height of 25.6 meters. The solution to this problem is best performed using the five term equation that you see listed below on the bottom of this page. Initially the ball has kinetic energy, and we'll assume initially the amount of potential energy is negligible, PEI is zero. There's work being done by a non-conservative force, air resistance, and we're told that the amount of work that it does is 10% of the original energy, in other words, 10% of the KEI. So, the left side of this equation is KEI plus W and C, where W and C is actually 10% of the KEI. Another way to reword this left side of the equation is to say it's 0.9 times KEI, where KEI is one half, MV, initial squared. On the right side of the equation, we have a PE final. The PE final is MGHF, where the HF is 25.6 meters, and we also have a KE final. If we wish to determine the speed of the ball, we need to find this KE final value, and then use it to calculate speed. So the problem would go something like this. We would begin on the left side of the equation calculating the KEI. We would go one half times 0.145 times 92.7 squared. That's one half MV initial squared. We get for that 623.01 joules, but we have to take 90% of that value, since 10% of the original energy is lost. That's where the W and C term comes in. So taking 90% of the 623.01 joules gives us 560.7122 joules. Now we know the left side of the equation. The right side of the equation has two terms, PE final and KE final. The PE final can be calculated as MGHF, where the HF is 25.6 meters, the M is 0.145 kilograms, and the G, of course, is 9.8 newtons per kilogram. Multiplying those three quantities together, we get 36.3776. The other term on the right side is the KEF. We wish to calculate it. And so what we do is we subtract 36.3776 from both sides of the equation, which leaves us with KEF equals 524.3346. Now that we know the final kinetic energy, we can use the equation KEF equals one half MVF squared to solve for the final velocity. We plug in 0.145 for the mass, and then we do proper algebra, finally taking the square root, and we get 85.0 meters per second.