# The path of a large stone fired from a torsion catapult can be modeled by y = -0.0054x^2 +...

## Question:

The path of a large stone fired from a torsion catapult can be modeled by {eq}y = -0.0054x^2 + 1.145x {/eq}, where {eq}x {/eq} is the distance the stone traveled (in yards) and y is the height of the stone (in yards).

1. find the distance the stone traveled.

2. find the maximum height of the stone.

## Maxima:

We need to know how to find the maxima of a function using differentiation in order to solve this question. For a function f(x), its maxima can be found by first differentiating f(x) by x and then equating it to 0, the value of x we will get, gives us the maxima of the function f(x)

Given: {eq}y = -0.0054x^2 + 1.145x {/eq}

1) The distance stone traveled can be found bu putting y=0,

\begin{align} y &= -0.0054x^2 + 1.145x \\ 0 &= -0.0054x^2 + 1.145x \\ x(-0.0054x + 1.145)&=0 \\ x&=\frac{1.145}{0.0054} \\ x&\approx 212 \end{align}

Hence, the distance traveled by stone is 212 yards.

2) Differentiating y with respect to x and equating it to zero, will give the value of x where the distance is maximum.

\begin{align} \frac{dy}{dx}&=\frac{d( -0.0054x^2 + 1.145x)}{dx} \\ \frac{dy}{dx}&=-0.0108x+1.145 \\ \mbox{ Equating it to zero} \\ -0.0108x+1.145 &=0 \\ x&=\frac{1.145}{0.0108} \\ x&\approx 106 \end{align}

Putting the value of x=106 in the equation, we will get the maximum height.

\begin{align} y&= -0.0054(106)^2 + 1.145(106) \\ y&=-0.0054(11236) + 1.145(106) \\ y&=-60.6744+121.37 \\ y&=60.6956 \\ y&\approx 61 \end{align}

Hence, the maximum distance is 61 yards.