Question

An object is launched directly in the air at a speed of 72 feet per second from a platform located 12 feet above the ground. The position of the object can be modeled using the function f(t)=−16t2+72t+12, where t is the time in seconds and f(t) is the height, in feet, of the object. What is the maximum height, in feet, that the object will reach?

Answer 1

Answer:

48

Step-by-step explanation:

For the quadratic ax^2 +bx +c, the axis of symmetry is x = -b/(2a). For the given quadratic, which defines a parabola opening downward, the axis of symmetry defines the time at which the maximum height is reached.

 t = -48/(2(-16)) = 1.5

Then the maximum height is ...

 f(1.5) = (-16·1.5 +48)1.5 +12 = (24·1.5) +12

 f(1.5) = 48

The maximum height the object will reach is 48 feet.

Answer 2

Answer:

93 feet

Step-by-step explanation:

i graphed it and found the maximum

or, if you're not allowed to use a calculator then find the vertex

a = -16, b = 72, c = 12

-b/2a will give you the x-value of 9/4

now plug 9/4 into the equation to get the value of 'y'

= -16(9/4)² + 72(9/4) + 12

= -81 + 162 + 12

= 93