| Sample Problem 2-7 | |
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In Figure 2-11, a pitcher tosses a baseball up along a y axis, with an initial speed of 12 m/s.
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| Fig. 2-11 A pitcher tosses a baseball straight up into the air. The equations of free fall apply for rising as well as for falling objects, provided any effects from the air can be neglected. |
(a) How long does the ball take to reach its maximum height?
Solution: One Key Idea here is that once the ball leaves the pitcher and before it returns to his hand, its acceleration is the free-fall acceleration a = -g. Because this is constant, Table 2-1 applies to the motion. A second Key Idea is that the velocity v at the maximum height must be 0. So, knowing v, a, and the initial velocity v0 = 12 m/s, and seeking t, we solve Equation 2-11, which contains those four variables. This yields
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(b) What is the ball's maximum height above its release point?
Solution: We can take the ball's release point to be y0 = 0. We can then write Equation 2-16 in y notation, set y - y0 = y and v = 0 (at the maximum height), and solve for y. We get
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(c) How long does the ball take to reach a point 5.0 m above its release point?
Solution: We know v0, a = -g, and displacement y - y0 = 5.0 m, and we want t, so we choose Equation 2-15. Rewriting it for y and setting y0 = 0 give us
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or
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If we temporarily omit the units (having noted that they are consistent), we can rewrite this as
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Solving this quadratic equation for t yields
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There are two such times! This is not really surprising because the ball passes twice through y = 5.0 m, once on the way up and once on the way down.
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