Question

Height of a Projected Ball An astronaut on the moon throws a baseball upward. The astronaut is 6 ft, 6 in. tall, and the initial velocity of the ball is 30 ft per sec. The height s of the ball in feet is given by the equations=-2.7t2+30t+6.5,where t is the number of seconds after the ball was thrown. (a) After how many seconds is the ball 12 ft above the moon's surface? Round to the nearest hundredth. (b) How many seconds will it take for the ball to hit the moon's surface? Round to the nearest hundredth.

Accepted Answer

Identify the given height function of the ball: $$s(t) = -2.7t^2$ + 30t + 6.5$$, where $$s(t) is $$the height in feet and $$t is $$the time in seconds. For part (a), set the height equal to 12 feet to find when the ball is 12 ft above the surface: $$-2.7t^2 + 30t + 6.5 = 12$. Rearrange the equation to standard quadratic form: -2.7t^2$ + 30t + 6.5 - 12 = 0$$, which simplifies to $$-2.7t^2 + 30t - 5.5 = 0$. Use the quadratic formula t$$ = \frac{-b \pm \sqrt{$$b^2 - 4ac$$}}{2a}$$ with a = -2.7$, b = 30$, and c = -5.5$ to solve for t$. This will give two possible times when the ball is at 12 ft. For part (b), find when the ball hits the moon's surface by setting s(t) = 0$: -2.7t^2$ + 30t + 6.5 = 0. $$Use the quadratic formula again with $$a = -2.7$$, $$b = 30$$, and $$c = 6.5 to $$solve for $$t. $$Choose the positive root as the time when the ball hits the surface.

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Key Concepts

Quadratic Functions and Their Graphs

Solving Quadratic Equations

Physical Interpretation of the Quadratic Model