Question

The function h(t) gives the height of a rock thrown off the Golden Gate Bridge in feet above the water at t seconds after it is thrown.

Question: How long after it’s thrown will the rock hit the water?

Answer 1

The rock will hit the water approximately 3.375 seconds after it's thrown.

To find when the rock hits the water, we need to determine the time (\(t\)) when the height
`(\(h(t)\))` is equal to zero.

The given function is
`\(h(t) = -16t^2 + 13t + 225\)`.

Setting
`\(h(t)\)` to zero and solving for t:

`\[ -16t^2 + 13t + 225 = 0 \]`

Now, we can use the quadratic formula to find t:

`\[ t = (-b \pm √(b^2 - 4ac))/(2a) \]`

In this case:

`\[ a = -16, \quad b = 13, \quad c = 225 \]`

`\[ t = (-13 \pm √(13^2 - 4(-16)(225)))/(2(-16)) \]`

`\[ t = (-13 \pm √(169 + 14400))/(-32) \]`

`\[ t = (-13 \pm √(14569))/(-32) \]`

`\[ t = (-13 \pm 121)/(-32) \]`

Now, we have two potential solutions for t:

1.
`\( t_1 = (-13 + 121)/(-32) \)`

2.
`\( t_2 = (-13 - 121)/(-32) \)`

Calculating these:

1.
`\( t_1 = (108)/(-32) \approx -3.375 \)` (ignoring the negative root since time cannot be negative in this context)

So, the rock will hit the water approximately 3.375 seconds after it's thrown.

The probable question may be: "The function
`\(h(t) = -16t^2 + 13t + 225\)` gives the height of a rock thrown off the Golden Gate Bridge in feet above the water at t seconds after it is thrown. How long after the rock thrown will hit the water?"