Question

The height, h, in feet of a rock above the ground is given by the equation h = -16t^2 + 24t + 16, where t is the time in seconds after the rock is thrown. At what time does the rock hit the ground?
Please show all work and not just the answer.

Answer 1

The rock will hit the ground at 2 seconds.

Explanation:

Given that the height, h, in feet of a rock above the ground is given by the equation $h=-16t^2+24t+16$, where t is the time in seconds after the rock is thrown.

We need to determine the time at which the rock will hit the ground.

To determine the time, let us equate the height h = 0 in the equation $h=-16t^2+24t+16$, we get,

$0=-16t^2+24t+16$

Switch sides, we have,

$-16 t^{2}+24 t+16=0$

Let us solve the equation using the quadratic formula.

Thus, we have,

$t=\frac{-24 \pm \sqrt{24^{2}-4(-16) 16}}{2(-16)}$

Simplifying, we get,

$t=\frac{-24 \pm \sqrt{576+1024}}{-32}$

$t=\frac{-24 \pm \sqrt{1600}}{-32}$

$t=\frac{-24 \pm 40}{-32}$

Hence, the two values of t are

$t=\frac{-24 + 40}{-32}$   and   $t=\frac{-24 -40}{-32}$

Simplifying the values, we get,

$t=\frac{16}{-32}$  and  $t=\frac{-64}{-32}$

Dividing, we get,

$t=-\frac{1}{2}$  and  $t=2$

Since, t cannot negative values, we have,

$t=2$

Thus, the rock will hit the ground at 2 seconds.