A baseball player standing on the ground throws a ball straight up. The ball leaves the player's hand with a speed of 15.5 m/s and is in the air for 3.30 s before hitting the ground. How high above the ground was the baseball player's hand when he released the ball?

v0=15.5

v1=0
v1=v0+at
Solve for t=(0-15.5)/(-9.81)=1.580 sec.
So it will return to the same elevation as it left off in 2*1.580=3.16 s.
Instead, it took 3.3-3.16=0.14 sec. more

Using the free-fall equation for H,
H=(1/2)gt^2, we have
H1=(1/2)g(1.580)^2
H2=(1/2)g(1.720)^2
Distance of "hand" from ground
= H2-H1
= 2.27m

To find the height above the ground at which the baseball player released the ball, we can use the kinematic equation:

h = h₀ + v₀t - (1/2)gt²

where:
h = height above the ground when the ball was released
h₀ = initial height, which is zero in this case since the baseball player is standing on the ground
v₀ = initial velocity, which is 15.5 m/s upwards
t = time, which is 3.30 s
g = acceleration due to gravity, which is approximately 9.8 m/s²

First, let's calculate the value of h using the equation:

h = 0 + (15.5 m/s)(3.30 s) - (1/2)(9.8 m/s²)(3.30 s)²

Simplifying the equation:

h = 0 + 51.15 m - (1/2)(9.8 m/s²)(10.89 s²)
h = 0 + 51.15 m - (53.682 m)
h ≈ -2.532 m

Since height cannot be negative in this context, we can conclude that the height above the ground at which the baseball player released the ball was approximately 2.532 meters.