45. A long jumper leaves the ground at an angle of 20° above the horizontal, at a speed of 11 m/sec. The height of the jumper can be modeled by h(x) = -0.046x2+ 0.364x, where h is the jumper's height in meters and x is the horizontal distance from the point of launch. a. At what horizontal distance from the point of launch does the maximum height occur? Round to 2 decimal places. b. What is the maximum height of the long jumper? Round to 2 decimal places. c.What is the length of the jump? Round to 1 decimal place.

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Asked Sep 24, 2019
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45. A long jumper leaves the ground at an angle of 20° above the horizontal, at a speed of 11 m/sec. The height of the jumper can be modeled by h(x) = -0.046x2+ 0.364x, where h is the jumper's height in meters and x is the horizontal distance from the point of launch. a. At what horizontal distance from the point of launch does the maximum height occur? Round to 2 decimal places. b. What is the maximum height of the long jumper? Round to 2 decimal places. c.What is the length of the jump? Round to 1 decimal place.

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Expert Answer

Step 1

Consider the given height as a function of horizontal length:

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h(x)=-0.046x 0.364x w wd

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Step 2

For maximum height,find the derivative of h(x),which in turn become zero.

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h'(x)=-0.046x 2x 0.364 h'(x)0 for maximum height 0.3643.96 0.092

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Step 3

Now find the maximum height by putting ...

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hma0.046 x3.962 +0.364 x 3.96 h 0.72 max

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Math

Calculus