CAn anyone solve this?

The big question is if this is a calculus class or not. Depending on that answer, I'd solve it differently. Presuming it's not, I'll solve this by putting it into vertex form:

h(t) = a(t - h) + k

When in this form, "k" is your answer as the maximum height:

h(t) = -16t² + 68t

I need t²'s coefficient to be 1, so divide both sides by -16:

-h(t)/16 = t² - (17/4)t

Now that we're in this format, we can complete the square on the right side by taking half of t's coefficient and squaring it. Then add that to both sides. So add 289/64 to both sides:

289/64 - h(t)/16 = t² - (17/4)t + 289/64

Now the right side can be factored as a perfect square trinomial:

289/64 - h(t)/16 = (t - 17/8)²

This tells us that the maximum height is reached at 17/8 seconds (2.125 seconds)

Now solve for h(t) again, keeping the binomial square in tact:

-h(t)/16 = (t - 17/8)² - 289/64
h(t) = -16(t - 17/8)² + 289/4

So 289/4 feet (72.25 ft) is the maximum height.

Yes. I can solve it using one of the kinematics formulas.

h ` (t) = 68 - 32 t = 0
t = 68 / 32 = 17 / 8 sec
h (17/8) = 68 x 17/8 - 16 x (17/8)²h (17/8) = [17/8 ] [ 68 - 16 x 17/8 ]
h (17/8) = [17/8 ] [ 34 ]
h (17/8) = 17²/4 = 72,25 ft

Sure. It's a 2nd degree equation, so the maximum will be at the vertex.

Using the formula for the vertex of ax^2 + bx + c form of 2nd degree equation, you want

to compute -b/2a

b = 68, a = -16 so thats -68/(2(-16)) = -68/-32 = 17/8 seconds

Then you plug in 17/8 second and you get the 72.25 ft answer that others have given.






b =

17/8 seconds so 289/4 feet or 72.25 feet

Differentiate and equate to zero.
h'(t) = 68 - 32t = 0
32t = 68
t = 68/32
t = 8.125 s is time to max. height .
Hence ma height is
h(8.125) = 68(8.125) - 16(8.125)^2
h(8.125) = 552.5 - 1056.24
h(8.125) = 503.75 ft.