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

Option c is correct

226 feet is the maximum height of the projectile.

Explanation:

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

then the axis of symmetry is given by:

`x = -(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 = -(48)/(2(-16)) = (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.