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(-(b)/(2a),f\left(-(b)/(2a)\right)\right)`

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

So,

`-(b)/(2a)=-(152)/(2(-16))`

`-(b)/(2a)=-(152)/(-32)`

`-(b)/(2a)=4.75`

So, x-coordinate of the vertex is 4.75.

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