Question

GRAVITY The height above the ground of a ball thrown up with a velocity of 96 feet per second from a height of 6 feet is 6+96t-16t^2 feet, where t is the time in seconds. According to this model, how high is the ball after 7 seconds? Explain.

Answer 1

Here the time function is h(t) = [6 + 96t - 16t^2] feet.

The initial height of the ball is 6 feet.  That's when t=0.  h(0)=[6+0-0] ft = 6 ft.

At t=7 sec, h(t) = [6 + 96t - 16t^2] feet becomes
       h(7 sec) = h(t) = [6 + 96(7) - 16(7)^2] feet       This produces a large negative number (-106 ft), which in theory indicates that the ball has fallen to earth and burrowed 106 feet into the soil.  Doesn't make sense.

Instead, let t=1 sec.    Then h(1 sec) = h(t) = [6 + 96(1) - 16(1)^2] feet
                                                                     =[6 + 96 -16] ft, or 86 ft.

One sec after the ball is thrown upward, it reaches a height of 86 feet.  It continues to rise, slowing down, until it finally stops for an instant and then begins to fall towards earth.