document.write( "Question 947649: 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
\n" ); document.write( "a. at what horizontal distance from the face of the cliff will the projectile strike the water?
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Algebra.Com's Answer #578338 by rothauserc(4718)\"\" \"About 
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we are given
\n" ); document.write( "h(x)= -32x^2/(50)^2+x+200, where x is time
\n" ); document.write( "let x1 be the axis of symmetry of this parabola that curves downward
\n" ); document.write( "x1 = -b/2a = (-1/2*(-32/50^2)) = 39.0625 seconds
\n" ); document.write( "now substitute in h(39.0625)
\n" ); document.write( "h(39.0625) = ((-32*(39.0625^2))/50^2)+39.0625+200 = 219.53125 feet max height
\n" ); document.write( "now the height of the cliff is 200 feet so 219.53125 - 200 = 19.53125 feet above cliff and remember that the projectile is launched at 45 degrees, therefore the horizontal distance to the max height is calculated
\n" ); document.write( "tan(45 degrees) = 19.53125 / d and
\n" ); document.write( "d = 19.53125 / tan(45 degrees) = 19.53125 / 1 = 19.53125 feet
\n" ); document.write( "d is the horizontal distance to the max height of the projectile, we have the following ratio
\n" ); document.write( "(39.0625 / 19.53125) = (170.023864174 / x2) where x2 is the horizontal distance to where the projectile strikes the water and 170.023864174 is the number of seconds when the projectile strikes the water (calculated by solving 0 = -32x^2/(50)^2+x+200)
\n" ); document.write( "x2 = (19.53125 * 170.023864174) / 39.0625 = 85.011932087 feet
\n" ); document.write( "The distance from the cliff to where the projectile strikes the water is 85.011932087 feet\r
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