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
a. at what horizontal distance from the face of the cliff will the projectile strike the water?
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! we are given
h(x)= -32x^2/(50)^2+x+200, where x is time
let x1 be the axis of symmetry of this parabola that curves downward
x1 = -b/2a = (-1/2*(-32/50^2)) = 39.0625 seconds
now substitute in h(39.0625)
h(39.0625) = ((-32*(39.0625^2))/50^2)+39.0625+200 = 219.53125 feet max height
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
tan(45 degrees) = 19.53125 / d and
d = 19.53125 / tan(45 degrees) = 19.53125 / 1 = 19.53125 feet
d is the horizontal distance to the max height of the projectile, we have the following ratio
(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)
x2 = (19.53125 * 170.023864174) / 39.0625 = 85.011932087 feet
The distance from the cliff to where the projectile strikes the water is 85.011932087 feet
|
|
|
