A rocket is launched from the top of a 50 foot cliff with an initial velocity of 120 feet per second. The height, h, of the rocket after t seconds is given by the equation h=-16t^2 + 120t+ 50. How long after the rocket is launched will it be 10 feet from the ground?

Well, if the height of the rocket is given by the equation h=-16t^2 + 120t+ 50, and we want to know when it will be 10 feet from the ground, we can set h equal to 10 and solve for t. So we have -16t^2 + 120t + 50 = 10. Let's simplify that equation:

-16t^2 + 120t + 40 = 0

Now, let's solve for t using the quadratic formula:

t = (-b ± sqrt(b^2 - 4ac))/(2a)

Plugging in the values for a, b, and c, we get:

t = (-(120) ± sqrt((120)^2 - 4(-16)(40)))/(2(-16))

Simplifying further:

t = (-120 ± sqrt(14400 + 2560))/(32)

t = (-120 ± sqrt(16960))/(32)

Now, calculating the square root of 16960:

t = (-120 ± 130.224)/(32)

We can break this down into two possibilities:

t = (-120 + 130.224)/(32) = 10.224/32 = 0.3195 seconds,

or

t = (-120 - 130.224)/(32) = -250.224/32 = -7.8205 seconds.

But wait, negative time? That doesn't make any sense! So, it looks like the rocket will be 10 feet from the ground after approximately 0.3195 seconds. And just in time to grab some popcorn and enjoy the show!

To find the time it takes for the rocket to be 10 feet from the ground, we need to solve the equation h = 10, where h is the height of the rocket after t seconds.

Substituting the given values into the equation, we have:

10 = -16t^2 + 120t + 50

Rearranging the equation, we get:

16t^2 - 120t - 40 = 0

Dividing the equation by 8, we have:

2t^2 - 15t - 5 = 0

Now, we can solve this quadratic equation using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 2, b = -15, and c = -5.

Substituting these values into the formula, we get:

t = (15 ± √((-15)^2 - 4(2)(-5))) / (2(2))

Simplifying the equation further:

t = (15 ± √(225 + 40)) / 4

t = (15 ± √265) / 4

Therefore, the two possible solutions for t are:

t = (15 + √265) / 4 or t = (15 - √265) / 4.

These are the two possible times, in seconds, when the rocket will be 10 feet from the ground after being launched.

To find the time when the rocket is 10 feet from the ground, we need to set the equation h = 10 and solve for t.

The equation for height is given by h = -16t^2 + 120t + 50.

Substituting 10 for h:
10 = -16t^2 + 120t + 50.

Let's rearrange the equation to bring it into standard quadratic form:
-16t^2 + 120t + 50 - 10 = 0,
-16t^2 + 120t + 40 = 0.

Next, let's divide the entire equation by 4 to simplify it:
-4t^2 + 30t + 10 = 0.

Since we cannot simplify this quadratic equation further, we can either use factoring, completing the square, or the quadratic formula to solve it.

The quadratic equation is of the form ax^2 + bx + c = 0, where in our case a = -4, b = 30, and c = 10.

One way to solve this equation is by factoring, but the factors may not always be easy to find. In this case, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a).

Plugging in the values, we get:
t = (-(30) ± √((30)^2 - 4(-4)(10))) / (2(-4)).

Simplifying further:
t = (-30 ± √(900 + 160)) / (-8).
t = (-30 ± √1060) / -8.

Now, we can approximate the value.

Using a calculator, we find that the square root of 1060 is approximately 32.62.

t = (-30 ± 32.62) / -8.

We have two possible solutions:
1. t = (-30 + 32.62) / -8,
2. t = (-30 - 32.62) / -8.

Simplifying each equation:
1. t = 2.62 / -8,
2. t = -62.62 / -8.

Calculating further:
1. t ≈ -0.3275,
2. t ≈ 7.8275.

Since time cannot be negative, the rocket will be 10 feet from the ground approximately 7.8275 seconds after it is launched.

just solve for t in

-16t^2 + 120t+ 50 = 10

probably the quadratic formula will be helpful.