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?
+36916 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)
