Afootball is punted from a height of 2.5 feet above the ground with an initial vertical velocity of 45 feet per second.
write an equation to model the height h in feet of the ball t seconds after it has been punted.
the football is caught at 5.5 feet above the ground. how long was the football in the air?

The standard kinematics equation is for an object projected vertically is:
H(t)=H0+v0(t)+(1/2)at^2
H0=height at time 0
v0(t)=vertical velocity at time 0
a=acceleration [equals -g for gravity]
H(t) height of projectile at time t.

We're given
H0=2.5'
v0=45 '/s
a=-32.2 '/s^2

so the kinematics equation is
H(t)=2.5+45(t)+(1/2)(-32.2)t^2= 2.5+45t-16.1t^2

Solve for H(t)=5.5 using the quadratic formula:
H(t)=5.5= 2.5+45t-16.1t^2
t=0.0683 s  [ on its way up ], or
t=2.727 s [caught on its way down]

Therefore the football was in the air for 2.727 seconds.

The 3rd quadrant as 3x+g5ti=56i: 92d-543