Since our projectile travels at a constant rate of 340 ft / sec in the horizontal direction, we have, using the formula D = RT, that our horizontal distance x = 340 t, giving
t = x / 340
Our above equation h(t) can now be converted into h(x), as followsh(x)=-16*(x/340)^2+340*x/340givingh(x)=-x^2/7225+xWhat is the ordered pair which constitutes the vertex of our parabolic equation h(x)?
At what distance from our firing point does our projectile reach maximum height?
What are the ordered pairs corresponding to the zeros of the above equation h(x)
(solutions to the equation -x^2/7225+x=0)?
How far does our projectile travel?The vertex of a parabola y=ax^2+bx+c is at the point x=-b/(2a). In our case we have a=-1/7225, and b=1, so the vertex is at
x=-1/(-2/7225) = 7225/2 = 3612.5ft. This is the distance from the firing point, where the projectile reaches its maximal height.To find the zeros of the function h(x), solve the equation
-x^2/7225 + x = 0
factor out x:
x(-x/7225 + 1) = 0
so x=0 or -x/7225 + 1=0, x/7225 = 1, x = 7225ft
The second number, x=7225ft is the maximal distance our projectile travels.

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