college algrebra word problem...problem in the details?

Question

a ball is shot up in the air and its height, h, above the ground in feet is given by the function h(x) = -16x^2 + 45x, where x is the number of seconds the ball has been in flight.graph this function and find the x-value for which the maximum height of the ball is attained.round to the hundredths place.

Answer 1

h(x)=-16x^2+45x
dh/dx=-32x+45
at maximum height dh/dx=0
so x=45/32
=1.41 sec.

Answer 2

To find maximums or minimums with a given function, take the derivative and set it equal to zero. This will show you everywhere the slope of the curve is zero!

h(x) = -16x^2 + 45x
h'(x) = -32x + 45 = 0

-32x + 45 = 0
32x = 45
x = 45/32

x = 1.40625 seconds
x ~ 1.41 seconds

Answer 3

h(x) = -16x^2 + 45x

Well, this is a dome shaped parabola, with vertex at the origin.

For h(max) you can either use differential calculus or simply complete the square:

1)Calculus:
differentiating the given eqn. d(h(x))/dx=-32x + 45
differentiating again d[d(h(x))/dx]/dx=-32
=>there is a maximum.
at:x=45/32=1.406~1.41 sec


2)Completing square:
h(x)=-(16x^2 - 45x)
=>h(x)=-16(x^2 - (45/16)x)
=>h(x)=-16(x^2 - (45/16)x +(45/32)^2 - (45/32)^2)
=>h(x)=-16[(x - 45/32)^2 - (45/32)^2]
=>h(x)= 16[(45/32)^2 - (x - 45/32)^2]
this is maximum if x=45/32~1.41 sec

Maximum ht. is 31.6 feet

Answer 4

h(x)=-16x^2+45x
h'=-32x+45
32x=45
x=45/32 sec for h-max
hmax=h(45/32)=-16*(45/32)^2+45*45/32
hmax=31.64 ft

Answer 5

Use differential calculus.

Answer 6

And after that, t = sqrt(2h/32.2)

so

1.41 = sqrt(2h/32.2)
1.99 = 2h/32.2
64.08 = 2h
height = 32.04 feet
:-)