An athlete whose event is the shot put releases the shot. When the shot is relea

Question

An athlete whose event is the shot put releases the shot. When the shot is released at an angle of 25 degree, its path can be modeled by the formula y = -0.01x 2 + 0.5x + 6.1 in which x is the shot's horizontal distance, in feet, and y is its height, in feet. This formula is shown by one of the graphs, (a) or (b), in the figure. Use the formula to answer the questions below. Use the formula to determine the shot's maximum distance. The maximum distance is approximately feet. (Round to the nearest tenth as needed.)

Explanation / Answer

your graph is 80 units long, , in increments of 10, and 40 units high, in increments of 10. Both x and y start at 0, but if you look at the start of these trajectories, you see they both start above zero, which makes sense, because a person throwing something will release it at the height of their hand, not at "y=zero". However, the shot WILL land at y=0, and this is the key to solving this problem. Replace y with 0 in the equation, and solve the quadratic equation, or use a competent calculator.... When solved for x, you obtain two solutions: x= -10.142 and change, and x= 60.1425 and change if you count off in units of 10 in the graph, you will see that (a) lands at around 40, and (b) lands at around 60.... If you were to trace b backward from the start point, you would see that it is also at zero at x=-10.142, your other answer... the answer is shot b, at 60.1 feet

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