7.) determine the initial velocity v so that its maximum height reached will be exactly 400 feet.

The derivative of height with respect to time is
  h'(t) = -32t +v
This will be zero when
  0 = -32t +v
  t = v/32

The height at this time is
  h(v/32) = -16(v/32)² + v(v/32) = (v²)(-1/(2·32) +1/32) = v²/64
Setting this to 400, we get
  400 = v²/64
  v = √(64·400) = 160

The initial velocity required so that the maximum height is exactly 400 feet is
  160 ft/s.