Question

Arthur drops a ball from a height of 81 feet above the ground. it’s height, , is given by the equation h=16ft^2+81, where t is the time in seconds. for which interval of the time is the height of the ball less than 17 feet

Answer 1

Answer:

The height of the ball is less than 17 feet for t > 2 seconds.

Step-by-step explanation:

The height of the ball after t seconds is given by the following equation:

$h(t) = -16t^{2} + 81$

For which interval of the time is the height of the ball less than 17 feet

This is t for which:

$h(t) < 17$

So

$-16t^{2} + 81 < 17$

$16t^{2} < -64$

Multiplying by -1

$16t^{2} > 64$

$t^{2} > \frac{64}{16}$

$t^{2} > 4$

$t > \sqrt{4}$

$t > 2$

The height of the ball is less than 17 feet for t > 2 seconds.