Question

An object is launched upwards from an initial height of 4 feet above ground, with an initial velocity of 48 feet per second. If the object's position at time x is given by f(x) = −16x2 + vt + s, where v is initial velocity and s is initial height, which equation can be used to find the maximum height of the object after x seconds?

Answer 1

The equation
`f(x)=-16(x-1.5)^2+40` is used to find the maximum height.

The maximum height of the object is 40 feet at x=1.5.

Explanation:

The vertex from of a parabola is

`f(x)=a(x-h)^2+k` ..... (1)

where, a is constant, (h,k) is vertex.

The object's position at time x is given by

`f(x)=-16x^2+vx+s`

where, v is initial velocity and s is initial height.

It is given that the initial height of the object is 4 feet and initial velocity is 48 feet per second.

Substitute v=48 and s=4 in the above function.

`f(x)=-16x^2+48x+4`

Rewrite the above equation in vertex form.

`f(x)=-16(x^2-3x)+4`

If an expression is
`x^2-bx`, then we need to add
`((b)/(2))^2` in it to make it perfect square.

In the parenthesis b=3,

`((3)/(2))^2=(1.5)^2`

Add and subtract (1.5)^2 in the parenthesis.

`f(x)=-16(x^2-3x+(1.5)^2-(1.5)^2)+4`

`f(x)=-16(x^2-3x+(1.5)^2)-16(-(1.5)^2)+4`

`f(x)=-16(x-1.5)^2+36+4`
`[\because (a-b)^2=a^2-2ab+b^2]`

`f(x)=-16(x-1.5)^2+40` .... (2)

The equation
`f(x)=-16(x-1.5)^2+40` is used to find the maximum height.

On comparing (1) and (2), we get

a=-16, h=1.5, k=40

Therefore, the maximum height of the object is 40 feet at x=1.5.