We have been given distance *d* (in feet) between the rock and the ground *t* seconds after it is thrown by the formula `d=-16t^2+3t+860` .

(a) First of all, we need to find the height of the rock after 6.5 seconds. To find the height of the rock after 6.5...

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We have been given distance *d* (in feet) between the rock and the ground *t* seconds after it is thrown by the formula `d=-16t^2+3t+860` .

(a) First of all, we need to find the height of the rock after 6.5 seconds. To find the height of the rock after 6.5 seconds, we need to replace *t* with 6.5 in the given formula and solve for *d*. Please go ahead and give it a shot before looking at the next step.

`d=-16(6.5)^2+3(6.5)+860`

`d=-16(42.25)+19.5+860`

`d=-676+19.5+860 `

`d=203.5`

Therefore, the rock will be 203.5 feet high after 6.5 seconds.

(b) For this part, we need to find the time when the rock would be at 590 feet. To find the time when the rock would be at 590 feet, we will replace *d* with 590 and solve for *t*. Please go ahead and give it a shot before looking at the steps ahead.

`590=-16t^2+3t+860`

`590-590=-16t^2+3t+860-590`

`-16t^2+3t+270=0`

Now we will use the quadratic formula to solve for *t*.

`t=(-b +- sqrt(b^2-4ac))/(2a)`

`t=(-3+-sqrt(3^2-4(-16)(270)))/(2*-16)`

`t=(-3+-3*sqrt(1921))/(-32)`

`t=(-3+3sqrt(1921))/(-32)` or `t=(-3-3sqrt(1921))/(-32) `

`t=-4.01523` or `t=4.20273`

Since time cannot be negative, the rock will be at 590 feet after 4.2 seconds.

(c) To find the time when the rock will hit the ground, we need to think about what the value of *d* will be when the rock hits the ground.

The value of *d* will be zero when the rock hits the ground, so we will replace *d* with 0 and solve for *t*. Please go ahead and give it a shot.

`-16t^2+3t+860=0`

We will again use the quadratic formula as we did for part b.

`t=(-3+-sqrt(3^2-4(-16)(860)))/(2*-16)`

`t=(-3+-sqrt(55049))/(-32)`

`t=(-3-sqrt(55049))/(-32)` or `t=(-3+3sqrt(1921))/(-32)`

`t = 7.42578` or

`t=-7.23828`

Since time cannot be negative, the rock will hit the ground after approximately 7.43 seconds.

(d) Now we need to find the maximum height of the rock. Can you recall how to find maxima for a parabola?

We can see that the leading coefficient of the parabola is negative (-16), which means our parabola will be a downward opening parabola and its maximum will be at the vertex of the parabola.

Are you familiar with the vertex formula of a parabola? If you cannot recall it, here it is:

The *t*-coordinate of the vertex is given by: `b/(-2a)` .

Please go ahead and substitute the values of *a* and *b* in the above formula.

`t=3/(-2*(-16))=3/32=0.09375`

Since the *x*-coordinate of the vertex is 0.09375, the rock will reach its maximum height at 0.09375 seconds.

The maximum height will be the value of *d* at *t* equal to 0.09375. Please go ahead and evaluate it using the formula given in the question.

`d=-16(0.09375)^2+3(0.09375)+860`

`d=-0.140625+0.28125+860 `

`d=860.140625 ~~ 860.14`

Therefore, the maximum height of the rock would be 860.14 feet.