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

Answer:

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.

Step-by-step 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 $(\frac{b}{2})^2$ in it to make it perfect square.

In the parenthesis b=3,

$(\frac{3}{2})^2=(1.5)^2$

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.

Answer 2

Answer:

f(x) = −16(x − 1.5)2 + 40

Step-by-step explanation:

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