SOLUTION: A projectile is fired at an inclination of 45 degrees to the horizontal, with a muzzle velocity of 100 feet per second. The height H of the projectile is modeled by {{{ H(x) =

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: A projectile is fired at an inclination of 45 degrees to the horizontal, with a muzzle velocity of 100 feet per second. The height H of the projectile is modeled by {{{ H(x) =       Log On


   

Question 1011428: A projectile is fired at an inclination of 45 degrees to the horizontal, with a muzzle velocity of 100 feet per second.
The height H of the projectile is modeled by
+H%28x%29+=+-32x%5E2%2F%28100%29%5E2%2Bx
Where x is the horizontal distance of the projectile from the firing point.
(A) At what horizontal distance from the firing point is the height of the projectile a maximum?
(B) Find the maximum height of the projectile.
(C) At what horizontal distance from the firing point will the projectile strike the ground?
(D) Using a graphing utility, graph the function H,
0(is equal to or less than) x (is equal to or less than) 350
(E) Using the graphing utility to verify the results obtained in parts B and C
(F) when the height of the projectile is 50 feet above the ground, how far has it traveled horizontally?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A projectile is fired at an inclination of 45 degrees to the horizontal, with a muzzle velocity of 100 feet per second.
The height H of the projectile is modeled by
+H%28x%29+=+-32x%5E2%2F%28100%29%5E2%2Bx
Where x is the horizontal distance of the projectile from the firing point.
(A) At what horizontal distance from the firing point is the height of the
projectile a maximum?
This is quadratic equation, the max will occur at the axis of symmetry; x=-b(2a)
Find a: -32/100^2 = .0064, b = 1
x = %28-1%29%2F%282%2A-.0032%29
x = +156.25 ft
:
(B) Find the maximum height of the projectile.
x = 156.25
H%28x%29+=+-32%28156.25%5E2%29%2F%28100%29%5E2%2B156.25
H(x) = 78.125 ft is the max ht
:
(C) At what horizontal distance from the firing point will the projectile strike the ground?
It will be twice the axis of symmetry
x = 2(156.25)
x = 312.5 ft from the origin when it hits the ground
:
(D) Using a graphing utility, graph the function H,
0(is equal to or less than) x (is equal to or less than) 350
Should look like this
+graph%28+300%2C+200%2C+-100%2C+350%2C+-10%2C+100%2C+%28-32x%5E2%2F100%5E2%29%2Bx%2C+50%29+

(E) Using the graphing utility to verify the results obtained in parts B and C
(F) when the height of the projectile is 50 feet above the ground, how far has it traveled horizontally?
h = 50
%28-32x%5E2%2F100%5E2%29%2Bx= 50
-.00032x%5E2+%2B+x+-+50+=+0
Solve this with the quadratic formula; a=-.0032, b=1, c=-50
Two solutions
x = 62.5 ft on the way up
and
x = 250 ft on the way down
(Green line on the graph is 50')