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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?

A82 feet
B190 feet
C226 feet
D250 feet

Verified Answer :

0821sonnicarter

The answer is C on edge

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.

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