A baseball player standing on the ground throws a ball straight up. The ball leaves the player's hand with a speed of 6.22 m/s and is in the air for 2.70 s before hitting the ground. How high above the ground was the baseball player's hand when he released the ball?

Answer 1

h = Hi + Vi t - 4.9 t^2

0 = Hi + 6.22(2.7) - 4.9 (2.7)^2