Question

Jake tosses a coin up in the air and lets it fall on the ground. The equation that

models the height (in feet) and time (in seconds) of the parabola is

h(t) = -16t2 + 24 + 6. Approximate the time at which the coin hits the

ground.

Answer 1

Answer:

$t = 1.71825$

Step-by-step explanation:

Given

$h(t) = -16t^2 + 24t + 6$

Required

When will the coin hit the ground

When the coin hits the ground, $h(t) = 0$

The expression $h(t) = -16t^2 + 24t + 6$ becomes

$0 = -16t^2 + 24t + 6$

Multiply through by -1

$16t^2 - 24t - 6 = 0$

Solve using quadratic formula

$t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}$

Where

$a = 16$

$b = -24$

$c = -6$

$t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}$

$t = \frac{-(-24)\±\sqrt{(-24)^2 - 4 *16 * -6}}{2 * 16}$

$t = \frac{24\±\sqrt{576 +384}}{32}$

$t = \frac{24\±\sqrt{960}}{32}$

$t = \frac{24\±30.984}{32}$

Split

$t = \frac{24+30.984}{32}$ or $t = \frac{24-30.984}{32}$

$t = \frac{54.984}{32}$ or $t = \frac{-6.984}{32}$

$t = 1.71825$ or $t = -0.21825$

But time can't be negative;

So:

Time to hit the ground is 1.71825 seconds