A model rocket is launches upward with an initial velocity of 160 feet per second. The height, in feet, of the rocket t seconds after the launch is given by h=-16t^2+160t. How many seconds after the launch will the rocket be 290 feet above the ground? Round to the nearest tenth of a second.

solve

290 = -16t^2 + 160t
16t^2 - 160t + 290 = 0

I don't think it factors, so use the formula to solve

To find the time it takes for the rocket to reach a height of 290 feet, we need to set the equation for the height equal to 290 and solve for t.

The equation for the height of the rocket is given by:
h = -16t^2 + 160t

Setting the equation equal to 290:
-16t^2 + 160t = 290

Next, we can rearrange the equation to have it equal to zero:
-16t^2 + 160t - 290 = 0

Now we can solve this quadratic equation for t. We can either use factoring, completing the square, or the quadratic formula.

Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)

For our quadratic equation, the coefficients are:
a = -16
b = 160
c = -290

Plugging these values into the quadratic formula, we have:
t = (-160 ± √(160^2 - 4(-16)(-290))) / (2(-16))

Simplifying the equation gives us:
t = (-160 ± √(25600 - 18560)) / (-32)
t = (-160 ± √(7040)) / (-32)
t = (-160 ± 83.92) / (-32)

Now we can solve for t:
t ≈ (-160 + 83.92) / -32 ≈ 2.24 seconds (rounded to the nearest tenth)

or

t ≈ (-160 - 83.92) / -32 ≈ 8.35 seconds (rounded to the nearest tenth)

Therefore, the rocket will be 290 feet above the ground approximately 2.24 seconds and 8.35 seconds after the launch.

To find the time it takes for the rocket to be 290 feet above the ground, we need to solve the equation -16t^2+160t = 290 for t.

1. Rewrite the equation: -16t^2 + 160t - 290 = 0

2. This is a quadratic equation in the form of at^2 + bt + c = 0, where a = -16, b = 160, and c = -290.

3. We can solve this quadratic equation using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

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

t = (-160 ± √(160^2 - 4(-16)(-290))) / (2(-16))

4. Simplify the equation:

t = (-160 ± √(25600 - 18560)) / (-32)

t = (-160 ± √7040) / (-32)

5. Calculate the square root of 7040:

√7040 ≈ 83.868

6. Plug in the values:

t = (-160 ± 83.868) / (-32)

7. Calculate both solutions:

t₁ = (-160 + 83.868) / (-32) ≈ 3.858

t₂ = (-160 - 83.868) / (-32) ≈ 9.142

8. Round the solutions to the nearest tenth of a second:

t₁ ≈ 3.9 seconds

t₂ ≈ 9.1 seconds

Therefore, the rocket will be approximately 290 feet above the ground after 3.9 seconds and 9.1 seconds.