Question

A rocket is launched from a tower. The height of the rocket, why in feet is related to the time after launch, X and seconds, by the given equation. Using this equation find out the time at which the rocket will reach its max, to the nearest 100th of a second y=-16x + 152x + 74

Answer 1

Given:

Consider the height of the rocket, in feet after x seconds of launch is

$y=-16x^2+152x+74$

To find:

The time at which the rocket will reach its max, to the nearest 100th of a second.

Solution:

We have,

$y=-16x^2+152x+74$

It is a quadratic polynomial with negative leading coefficient. So, it is a downward parabola.

Vertex of a downward parabola is the point of maxima.

To find the time at which the rocket will reach its max, we need to find the x-coordinate of the vertex.

If a quadratic function is $f(x)=ax^2+bx+c$, then the vertex is

$Vertex=\left(-\dfrac{b}{2a},f\left(-\dfrac{b}{2a}\right)\right)$

Here, $a=-16,b=152,c=74$.

So,

$-\dfrac{b}{2a}=-\dfrac{152}{2(-16)}$

$-\dfrac{b}{2a}=-\dfrac{152}{-32}$

$-\dfrac{b}{2a}=4.75$

So, x-coordinate of the vertex is 4.75.

Therefore, the rocket will reach its max at 4.75 second.