Question

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

Answer 1

totally different way by using kinematic eqns

height = init vel * t + 1/2 * accel * t^2

and

final vel ^2 = init vel ^2 + 2 * accel * height

given h(t) = vt - 16t^2

accel = -16*2 = -32

at max height=400, final vel = 0

0 = v^2 + 2 * (-32) * 400

v^2 = 25600

v = 160 feet per second

Answer 2

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.