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3 years ago

10

A projectile is fired from a cliff 200 above the water at an inclination of 45 degree to the horizontal, with a muzzle velocity

of 50 feet per second. The height is given by h(x)= -32x^2/(50)^2+x+200
a. At what horizontal distance from the face of the cliff will the projectile strike the water?

1 answer:

iVinArrow [24]3 years ago

3 0

<h2>Horizontal distance when projectile strikes water surface is 170 .02 feet</h2>

Step-by-step explanation:

The height is given by

$h(x)=-\frac{32}{50^2}x^2+x+200$

We need to find when it strikes the water surface,

At water surface h = 0 feet

Substituting

$h(x)=-\frac{32}{50^2}x^2+x+200=0$

Solving this quadratic equation to get x.

-32x² + 2500x + 500000 = 0

32x² - 2500x - 500000 = 0

x = 170 .02 feet or x = -91.90 feet

Negative x is not possible

Horizontal distance when projectile strikes water surface = 170 .02 feet

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Step-by-step explanation:

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