Answer:
39 feet
Step-by-step explanation:
In this problem, the height of the football at time t is modelled by the equation:
[tex]h(t)=-16t^2+vt+s[/tex]
where:
s = 3 ft is the initial height of the ball
v = 48 ft/s is the initial vertical velocity of the ball
[tex]-32 ft/s^2[/tex] is the acceleration due to gravity (downward)
Substituting these values, we can rewrite the expression as
[tex]h(t)=-16t^2+48t+3[/tex]
Here we want to find the maximum height reached by the ball.
This is equivalent to find the maximum of the function h(t): the maximum of a function can be found requiring that the first derivative of the function is zero, so
[tex]h'(t)=0[/tex]
Calculating the derivative of h(t), we find:
[tex]h'(t)=-32 t+48[/tex]
And imposing it equal to zero, we find the time t at which this occurs:
[tex]0=-32t+48\\t=-\frac{48}{-32}=1.5 s[/tex]
And substituting back into h(t), we can find the maximum height of the ball:
[tex]h(1.5)=-16\cdot (1.5)^2 + 48\cdot 1.5 +3=39 ft[/tex]
