Can somebody please help me with this problem?

h(t) = -16t^2+64t+29 

The math way:

Notice that this is a downward Parabola (why?). Therefore it's max is its vertex. So what is the vertex of a parabola?

Thy physics/kinematics way:

Given your height as a function of time, what is its velocity as a function of time?

The max height occurs when it's velocity is zero. So just set v(t) =  0 and you find the time. Replace this time into your height function.

Done!

Show your steps and tentative answers here and we will tell you if it's okay.

Don't forget to vote me best answer for being the first to correctly walk you through without spoiling the answer. That way it gives you a chance to work at it and to get good at it.

You are given the equation:

h(t) = -16t² + 64t + 29

And asked to find the maximum height.

Since this is college algebra, I'll show you how this can be done without using calculus.  You want to get this into vertex form:

h(t) = a(t - h)² + k

The vertex is the point (h, k) which tells you the time it takes to get to the maximum (h) and what the maximum height is (k), which is what you are being asked for.

To do that we need to get the right side to be in the form of (t² + bt).  We will subtract 29 from both sides then divide both sides by -16:

h(t) = -16t² + 64t + 29

h(t) - 29 = -16t² + 64t

-[h(t) - 29] / 16 = t² - 4t

Now we complete the square by adding 4 to both sides:

-[h(t) - 29] / 16 + 4 = t² - 4t + 4

The right side can now be factored:

-[h(t) - 29] / 16 + 4 = (t - 2)²

Solve for h(t) again.  Subtract 4 from both sides multiply both sides by -16, then add 29 to both sides:

-[h(t) - 29] / 16 = (t - 2)² - 4

h(t) - 29 = -16(t - 2)² + 64

h(t) = -16(t - 2)² + 93

The maximum height is 93 feet reached at 2 seconds after the arrow's launch.