QuestionA golf ball is hit at 130 ft/s at $45^{\circ}$ . Find the distance to max height using $h(x)=\frac{-32 x^{2}}{130^{2}}+x$ .

Studdy Solution

STEP 1

Assumptions1. The initial velocity of the golf ball is130 feet per second. . The inclination of the hit is $45^{\circ}$ to the horizontal.
3. The height $h$ of the golf ball is given by the function $h(x)=\frac{-32 x^{}}{130^{}}+x$ .
4. We need to find the distance the ball travels before it reaches a maximum height.

STEP 2

The maximum height is achieved when the derivative of the height function is zero, because the derivative represents the rate of change of the height. When the height is at a maximum, the rate of change is zero (the height is not increasing or decreasing at that point). So, we need to find the derivative of the height function.
$h'(x) = \frac{d}{dx} \left(\frac{-32 x^{2}}{130^{2}}+x\right)$

STEP 3

To differentiate the function, we apply the power rule, which states that the derivative of $x^n$ is $n*x^{n-1}$ .
$h'(x) = \frac{d}{dx} \left(\frac{-32 x^{2}}{130^{2}}\right) + \frac{d}{dx} (x)$

STEP 4

Differentiate each term separately.
$h'(x) = \frac{-64x}{130^{2}} +1$

STEP 5

Set the derivative equal to zero and solve for $x$ to find the distance the ball travels before it reaches a maximum height.
$\frac{-64x}{130^{2}} +1 =0$

STEP 6

Rearrange the equation to isolate $x$ .
$\frac{-64x}{130^{2}} = -1$

STEP 7

Multiply both sides by $-1$ .
$\frac{64x}{130^{2}} =1$

STEP 8

Multiply both sides by $130^{2}$ .
$64x =130^{2}$

STEP 9

Divide both sides by64 to solve for $x$ .
$x = \frac{130^{2}}{64}$

STEP 10

Calculate the value of $x$ .
$x = \frac{130^{2}}{64} \approx262.81$ The ball travels approximately263 feet (rounded to the nearest whole number) before it reaches a maximum height.

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