Expert: Stephen King - 8/11/2004

Question
    Hi there Stephen.  How are you doing today?  This problem is giving me trouble.  I was wondering if you can help me.  Here it is:

    The height, in feet, above the ground of a ball thrown straight up with an initial velocity of 160ft/sec was given by using the formula:

   h(t) = -162t2 + 160t + 4

    where t stood for the number of seconds after the ball had been thrown.  Fine the values of t for which the height of the ball was 260 feet.

Answer
Hey Damon.

Ok, I'm not sure which math you're taking, but using Calculus will yield that there is no solution.  The maximum height that the ball goes is approx 43.5 ft.  So the ball never got to the height of 260 feet.

If you're not in Calculus, think of it like this:  plug in values in for t, from 1 on up.  You'll see that because of the t^2 value, and the negative sign in front of the 162 that the ball will never go that high.

You can also graph the function on a graphing calculator and see that the heighest the ball goes is nowhere near 260 feet.

If this doesn't help, let me know.

Steve