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?

Answer 1

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