Question

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 50 ft per sec. The height s of the ball in feet is given by the equation s =-27t2 + 50t+ 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 18 ft above the moon's surface?

Answer 1

Given, S = -2.7 t^2 + 50 t + 6.5 (a) 18 = -2.7 t^2 + 20 t + 6.5 0 = -2.7 t^2 + 50 t + 6.5 - 18 = > -2.7 t^2 + 50 t - 11.5 = 0 t = -50 plusminus squareroot 50^2 - 4 (-2.7) (-11.5)/2 (-2.7) t = -50 plusminus squareroot 2500 - 124.2/-5.4 t = -50 plusminus 48.7421/-5.4 t = -50 + 48.7421/-5.4 t = 0.23292 t = 0.233 sec t = -50 - 48.7421/-5.4 t = 1802855 t = 18.3 sec