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?
h = Hi + Vi t - 4.9 t^2
0 = Hi + 6.22(2.7) - 4.9 (2.7)^2
To find the height above the ground where the baseball player's hand released the ball, we can use the kinematic equation for vertical motion:
h = (vi * t) - (1/2 * g * t^2)
Where:
h is the height above the ground
vi is the initial velocity (speed) of the ball
t is the time the ball is in the air
g is the acceleration due to gravity (approximately 9.8 m/s^2)
We are given the initial velocity (vi = 6.22 m/s) and the time in the air (t = 2.70 s).
Using the equation, we can plug in the given values:
h = (6.22 * 2.70) - (1/2 * 9.8 * 2.70^2)
Simplifying further:
h = 16.794 - (1/2 * 9.8 * 7.29)
h = 16.794 - (44.97)
h = -28.176
The negative sign indicates that the height is below the reference point (in this case, the ground), so we disregard it. The height above the ground where the baseball player's hand released the ball is approximately 28.176 meters.