Question 115783
If the position is denoted as h(t), then the instantaneous velocity at any given time is simply the first derivative, h'(t), while the acceleration is the second derivative, h''(t).   

If you differentiate the given position function twice, you will obtain a value of acceleration equal to -32 ft/sec^2, where I have chosen the units arbitrarily. The gravitational force (mg) acts on the projectile as it moves upwards. Ignoring air resistance, this is the only force acting on the projectile. Since the acceleration vector points radially inward, the projectile slows down as it rises. At the highest point, all of the kinetic energy of the projectile becomes potential energy, e.g. KE=PE.

To find the velocity of the projectile at it's highest point, take the first derivative:

h'(t)=-32t+208

Set h'(t) equal to 0: -32t+208=0

                         t=9.625 s (Answer)

The projectile reaches its maximum height after 9.625 s

The maximum height is given as follows:

h(6.5)= -16(9.625)(9.625)+308(9.625)=-1482.25+2964.5=1482.25 ft (Answer)


Denote the side length of a square by s

Then the total perimeter P of the square is 4s

The total area A of the square is s^2

We are told that the area is 12 more than the perimeter.

Therefore, s^2=4s+12

Rewrite as a quadratic equation:

s^2-4s-12=0

Factor:

(s+2)(s-6)=0

Only the positive root is significant. Therefore, s=6 (Answer)