Answer:
Option c is correct
226 feet is the maximum height of the projectile.
Step-by-step explanation:
A quadratic equation [tex]y=ax^2+bx+c[/tex], ....[1]
then the axis of symmetry is given by:
[tex]x = -\frac{b}{2a}[/tex]
As per the statement:
The path of the projectile is modeled using the equation :
[tex]h(t) = -16t^2+48t+190[/tex] ....[2]
where, h(t) is the height after t time.
On comparing with [1] we have;
a = -16 and b = 48
then;
[tex]t = -\frac{48}{2(-16)} = \frac{48}{32} = 1.5[/tex] sec
Substitute this in [2] we have;
[tex]h(1.5) = -16(1.5)^2+48(1.5)+190[/tex]
⇒[tex]h(1.5) = -36+72+190[/tex]
Simplify:
[tex]h(1.5) = 226 ft[/tex]
Therefore, the maximum height of the projectile at 1.5 sec is, 226 feet.
