Question

can someone show me the steps to this question

Answer 1

A projectile is thrown upward so that its distance above the ground after t seconds is h = –16t^2 + 672t. After how many seconds does it reach its maximum height?

Answer 2

Take the function h, and find its derivative (rate of change). When you do that, make the derivative equal to zero - which means you will locate the maximum point.

You will get a result in terms of t=c where c is a constant. You can then replace that in the initial function h, and get the maximum height. But your interest is only t=c.

Answer 3

can you please show me?

Answer 4

You can also just set the equation (as-is) to 0 and solve for t (a quadratic equation). You'll get two values for t, one of which will be 0 (seconds). The larger value will be when it hits the ground. Half of that larger value will be the time at which it's at its maximum height time. Once you have that you can just plug that time into the equation and solve for h.

Answer 5

I think it is much simpler to do Shane's version if you are not comfortable with calculus. If you need the calculus version though, we can guide you.

Answer 6

To explain that better:
\[h = –16t^2 + 672t\]\[ 0 = –16t^2 + 672t\]Solve the quadratic to get:\[t=0s, t=42s\]\[\frac{42s}{2s}=21\]\[h= –16(21s)^2 + 672(21s)=?\]

Answer 7

21s

Answer 8

Yes, @21 seconds it will be at it's maximum height.

Answer 9

Oh..I thought it also wanted to know what the max height was. That's what plugging 21s back into the original equation will get you.

Answer 10

what confused?

Answer 11

I showed you how to solve it for the time at which the height is at max and what the max height will be. You can ignore the second part of what I posted above. The answer is still 21 seconds.

Using the other method (calculus) you take the derivative of:\[h = –16t^2 + 672t\]\[h'=-32t+672\]
Set that equation to 0 and solve for t:\[0=-32t+672\]\[t=21seconds\]
Same answer either way.