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QuestionA golf ball is hit at 130 ft/s at 45∘45^{\circ}. Height is given by h(x)=−32x21302+xh(x)=\frac{-32 x^{2}}{130^{2}}+x. Find distance and max height.

Studdy Solution

STEP 1

Assumptions1. The initial velocity of the golf ball is130 feet per second. The inclination to the horizontal is 45∘45^{\circ}
3. The height hh of the golf ball is given by the function h(x)=−32x130+xh(x)=\frac{-32 x^{}}{130^{}}+x4. xx is the horizontal distance that the golf ball has traveled

STEP 2

The golf ball will reach its maximum horizontal distance when it hits the ground, i.e., when h(x)=0h(x) =0. So, we set the height function equal to zero and solve for xx.
0=−32x21302+x0=\frac{-32 x^{2}}{130^{2}}+x

STEP 3

To make the equation easier to solve, we can multiply each term by 1302130^2.
0=−32x2+1302x0=-32x^2+130^2x

STEP 4

This is a quadratic equation in the form of ax2+bx+c=0ax^2+bx+c=0, where a=−32a=-32, b=1302b=130^2, and c=0c=0. We can solve this quadratic equation for xx using the quadratic formula, x=−b±b2−4ac2ax=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.

STEP 5

Substitute aa, bb, and cc into the quadratic formula to solve for xx.
x=−1302±(1302)2−4∗(−32)∗02∗(−32)x=\frac{-130^2\pm\sqrt{(130^2)^2-4*(-32)*0}}{2*(-32)}

STEP 6

Since the term under the square root, b2−4acb^2-4ac, is zero, the equation simplifies tox=−13022∗(−32)x=\frac{-130^2}{2*(-32)}

STEP 7

Calculate the value of xx.
x=−13022∗(−32)=1302/64≈263x=\frac{-130^2}{2*(-32)} =130^2/64 \approx263To the nearest25 feet, the ball travels263 feet.

STEP 8

The maximum height of the ball is reached when the derivative of the height function, h′(x)h'(x), is zero. The derivative of h(x)h(x) ish′(x)=−64x1302+1h'(x)=\frac{-64x}{130^2}+1Setting h′(x)=0h'(x)=0 and solving for xx gives the horizontal distance at which the maximum height is reached.

STEP 9

Set h′(x)h'(x) equal to zero and solve for xx.
=−64x1302+=\frac{-64x}{130^2}+

STEP 10

olve the equation for xx.
x=130264=130/2=65x=\frac{130^2}{64} =130/2 =65

STEP 11

Substitute x=65x=65 into the height function h(x)h(x) to find the maximum height.
h(65)=−32(65)130+65h(65)=\frac{-32 (65)^{}}{130^{}}+65

STEP 12

Calculate the maximum height.
h(65)=−32(65)21302+65≈32.50h(65)=\frac{-32 (65)^{2}}{130^{2}}+65 \approx32.50The maximum height is approximately32.50 feet (rounded to two decimal places).

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