If a baseball pitch leaves the pitcher's hand horizontally at a velocity of 150 km/h by what percent will the pull of gravity change the magnitude of the velocity when the ball reaches the batter, 18 m away? for this estimate, ignore air resistance and spin on the ball.

0.52% First, let's convert that speed into m/s. 150 km/h * 1000 m/km / 3600 s/h = 41.667 m/s Now let's see how much time gravity has to work on the ball. Divide the distance by the speed. 18 m / 41.667 m/s = 0.431996544 s Now multiply that time by the gravitational acceleration to see what the vertical component to the ball's speed that gravity adds. 0.431996544 s * 9.8 m/s^2 = 4.233566131 m/s Use the pythagorean theorem to get the new velocity of the ball. sqrt (41.667^2 + 4.234^2) = 41.882 m/s Finally, let's see what the difference is (41.882 - 41.667) / 41.667 = 0.005159959 = 0.5159959% Rounding to 2 figures, gives 0.52%