SOLUTION: 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
Question 1152706: 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 equation s equals -2.7t^2 + 30t + 6.5, where t is the number of seconds after the ball was thrown. Complete parts a and b.
a. After how many seconds is the ball 12 ft above the moon's surface? Found 2 solutions by ikleyn, Alan3354:Answer by ikleyn(52778) (Show Source): You can put this solution on YOUR website! .
Solve the quadratic equation
-2.7t^2 + 30t + 6.5 = 12
using the quadratic formula.
The two solutions will give you two time moments.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website! s equals -2.7t^2 + 30t + 6.5, where t is the number of seconds after the ball was thrown. Complete parts a and b.
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a. After how many seconds is the ball 12 ft above the moon's surface?
s(t) = -2.7t^2 + 30t + 6.5 = 12
2.7t^2 - 30t + 5.5 = 0
Solve for t
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The smaller value is ascending.
The larger is descending.