Question

A football 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?

Answer 1

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.