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An astronaut on the moon throws a baseball upward. the astronaut is 6ft, 6 in. tall and the initial velocity of the ball is 30 ft per second. the height s of the ball in feet is given by the equation 2.7 t^2 + 50 + 6.5 where t is the number of seconds after the ball is thrown. After how many seconds is the ball 12 ft above the moons surface?

 Feb 7, 2020
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The height s of the ball in feet is given by the equation 2.7 t^2 + 50t + 6.5    So equate this to 12 ft and solve for 't'

2.7 t^2 + 50t + 6.5 = 12

2.7 t^2 + 50t  -5.5=0       Now use quadratic formula to find t =0.109354 sec   and  −18.6279 (throw out)

 Feb 7, 2020

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