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?

The equation 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

        ..... (1)

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

The object's position at time x is given by

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.

Rewrite the above equation in vertex form.

If an expression is , then we need to add in it to make it perfect square.

In the parenthesis b=3,

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

           .... (2)

The equation 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.

- 70

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