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

`t = 1.71825`

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 = (-b\±√(b^2 - 4ac))/(2a)`

Where

`a = 16`

`b = -24`

`c = -6`

`t = (-b\±√(b^2 - 4ac))/(2a)`

`t = (-(-24)\±√((-24)^2 - 4 *16 * -6))/(2 * 16)`

`t = (24\±√(576 +384))/(32)`

`t = (24\±√(960))/(32)`

`t = (24\±30.984)/(32)`

Split

`t = (24+30.984)/(32)` or
`t = (24-30.984)/(32)`

`t = (54.984)/(32)` or
`t = (-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