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(1 point)
A hot air balloon is launched from the ocean beach. As it moves inland, it rises and then falls, having a path given by
p(x)= - 2x^2/2500 +0.7x where x is in metres. The land rises at a rate of 2 vertical metres for each 20 horizontal metres. See the
diagram



a. At what horizontal distance does the balloon reach maximum height above sea level?
b. What is the maximum height of the balloon above seal level?
c. What is the maximum height of the balloon above ground level?
m and
d. At what horizontal distance is the balloon 50 m above the ground? List the smaller value first.

1 Point A Hot Air Balloon Is Launched From The Ocean Beach As It Moves Inland It Rises And Then Falls Having A Path Given By Px 2x22500 07x Where X Is In Metres class=

Sagot :

MrRoyal

The launching of the hot air balloon is from the ocean beach is an illustration of a projectile motion

The horizontal distance which it attains a maximum height

The function is given as:

[tex]p(x) =-\frac{2}{2500}x^2 + 0.7x[/tex]

Differentiate

[tex]p'(x) =-\frac{4}{2500}x + 0.7[/tex]

Set to 0

[tex]-\frac{4}{2500}x + 0.7 = 0[/tex]

Subtract 0.7 from both sides

[tex]-\frac{4}{2500}x =- 0.7[/tex]

Multiply through by -2500/4

[tex]x =437.5[/tex]

Hence, the horizontal distance which a maximum height is attained is 437.5 meters

The maximum height above the sea level

In (a), we have:

[tex]x =437.5[/tex]

Substitute [tex]x =437.5[/tex] in p(x)

[tex]p(437.5) =-\frac{2}{2500} * 437.5^2 + 0.7 * 437.5[/tex]

Evaluate

[tex]p(437.5) =153.125[/tex]

Hence, the maximum height above the sea level is 153.125 metres

The maximum height above the ground level

In (b), the maximum height above the sea level is 153.125 metres

The ratio of the height on land and sea is given as:

Land : Sea = 2 m : 20 m

For the maximum height, we have:

Land : 153.125 m= 2 m : 20 m

Express as fraction

Land / 153.125 m = 2 /20

Multiply both sides by 153.125

Land = 15.3125

Approximate

Land = 15.3

Hence, the maximum height above the ground level is 15.3 meters

The horizontal distance where the balloon 50 m above the ground

We have:

[tex]p(x) =-\frac{2}{2500}x^2 + 0.7x[/tex]

Substitute 50 for P(x)

[tex]-\frac{2}{2500}x^2 + 0.7x = 50[/tex]

Using a graphing calculator, we have:

x = 78.465 and 796.535

Hence, the horizontal distances where the balloon 50 m above the ground are 78.465 and 796.535 meters

Read more about projectile motion at:

https://brainly.com/question/1130127

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