What is the maximum height of the ball if it is thrown from a 12 ft. shelf at a speed of 32 ft/s. Use f(x)=-2x^2+2.

Question

A ball is thrown from the top of a shelf that is 12 feet tall at a speed of 32 feet per second which will yeild the equation f(x)=-2x^2+2. What will be the ball's maximum height?

Answer 1

I assume the ball is thrown directly upward at 32 ft/sec, or at least that the vertical component of the ball's initial motion is 32 ft/sec upward.

If so, given -32 ft/sec/sec as acceleration of gravity, we can determine that the ball will reach its highest point in 1 second, at which time its velocity will be 0.  From then on, the ball will fall.  During that second, the ball will travel upward 16 feet.  (Multiply time - 1 second - by mean upward velocity - (32+0)/2).  The ball was 12 feet off the floor when it was thrown, so maximum height = 12 + 16 = 28.