Question

What is the maximum height of the projectile? 82 feet 190 feet 226 feet 250 feet.

Answer 1

The maximum height of the projectile will be 226 feet that is, as per question, the option c.

A quadratic equation , y = ax²+bx+c  (equation 1)

given, axis of symmetry is

x = $\frac{b}{2a}$

As per the question:

The path of the projectile is modeled using the equation :

h(t) = -16t²+48t+190  (equation 2)

h(t) = height after t time.

comparing equation 1 and equation 2, we get

a = -16 and b = 48

further, t = $\frac{48}{2(-16)}$

= $\frac{48}{32}$

= 1.5sec

Substituting the found value in equation 2, we get

h(1.5) = -16(1.5)²+48(1.5)+190

h(1.5) = -36+72+190

h(1.5) = 226ft.

Thus, the maximum height of the projectile at 1.5 sec is, 226 feet.

Note that the full question is:

A projectile with an initial velocity of 48 feet per second is launched from a building 190 feet tall. The path of the projectile is modeled using the equation h(t) = –16t2 + 48t + 190.

What is the maximum height of the projectile?

A. 82 feet

B. 190 feet

C. 226 feet

D. 250 feet

To learn more about velocity: brainly.com/question/25749514

#SPJ4