Question

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

Answer 1

The answer is C on edge

Answer 2

Answer:

Option c is correct

226 feet  is the maximum height of the projectile.

Step-by-step explanation:

A quadratic equation $y=ax^2+bx+c$,   ....[1]

then the axis of symmetry is given by:

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

As per the statement:

The path of the projectile is modeled using the equation :

$h(t) = -16t^2+48t+190$            ....[2]

where, h(t) is the height after t time.

On comparing with [1] we have;

a = -16 and b = 48

then;

$t = -\frac{48}{2(-16)} = \frac{48}{32} = 1.5$ sec

Substitute this in [2]  we have;

$h(1.5) = -16(1.5)^2+48(1.5)+190$  

⇒$h(1.5) = -36+72+190$  

Simplify:

$h(1.5) = 226 ft$

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