Abaseball team is practicing throwing balls vertically upward to test their throwing arms. a player manages to throw a ball that reaches a maximum altitude of 10 m above the launch point, before falling back down. the acceleration due to gravity is 9.80 m/s 2 . (a) with what initial speed was the ball thrown? (b) how long did it take for the ball to reach its maximum altitude?

a) 14.0 m/s b) 1.43 s Ignoring air resistance, the ball falling from 10 meters will be exactly mirror the ball climbing 10 meters. Since the formula for how far an object travels under constant acceleration is d = 0.5 AT^2 all we need to do is substitute the known values and solve for T. So 10 m = 0.5 * 9.8 m/s^2 T^2 10 m = 4.9 m/s^2 T^2 2.040816327 s^2 = T^2 1.428571429 s = T So we now know the ball took 1.43 seconds to reach its maximum altitude. The initial speed the ball was thrown at will be the product of the time and gravitational acceleration. So 1.43 s * 9.80 m/s^2 = 14.0 m/s