A model rocket is launched upward with an initial velocity of 200 feet per second. The height, in feet, of the rocket t seconds after the launch is given by h = −16t2 + 200t. How many seconds after the launch will the rocket be 380 feet above the ground? Round to the nearest tenth of a second. (Enter your answers as a comma-separated list.)

Question

A model rocket is launched upward with an initial velocity of 200 feet per second. The height, in feet, of the rocket t seconds after the launch is given by h = −16t2 + 200t. How many seconds after the launch will the rocket be 380 feet above the ground? Round to the nearest tenth of a second. (Enter your answers as a comma-separated list.)

Answer 1

Well, let's solve this puzzle and launch ourselves into the math-mosphere!

We need to find the time when the height of the rocket is 380 feet. So, we can set up the equation:

380 = -16t^2 + 200t

Let's bring everything to one side and create a quadratic equation:

16t^2 - 200t + 380 = 0

Now, let me summon my trusty clown calculator to give us the solution:

The time, my friend, will be approximately 4.4 seconds.

So, after approximately 4.4 seconds, the rocket will be 380 feet above the ground. Rocket science, huh?

Answer 2

To find the time when the rocket will be 380 feet above the ground, we need to solve the equation h = 380.

Let's substitute h = 380 in the given equation:
380 = -16t^2 + 200t

To solve the equation, rearrange it to quadratic form:
-16t^2 + 200t - 380 = 0

Now, let's solve this quadratic equation. We can do this by factoring or using the quadratic formula.

For this example, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a

In our equation, a = -16, b = 200, and c = -380. Plugging these values into the quadratic formula, we get:
t = (-200 ± √(200^2 - 4(-16)(-380))) / 2(-16)

Simplify the equation:
t = (-200 ± √(40000 - 24320)) / -32
t = (-200 ± √(15680)) / -32
t = (-200 ± 125.200) / -32

Divide numerator by -32:
t = (200 ± 125.200) / 32

Now, calculate the two possible values of t:
t1 = (200 + 125.200) / 32 ≈ 9.786
t2 = (200 - 125.200) / 32 ≈ 2.928

Therefore, the rocket will be approximately 380 feet above the ground after 9.8 seconds and 2.9 seconds.

Answer 3

To find the time at which the rocket will be 380 feet above the ground, we need to solve the equation h = -16t^2 + 200t for t when h = 380.

The equation is given as h = -16t^2 + 200t, where h represents the height of the rocket in feet and t represents the time in seconds.

Setting h to 380, we have:
380 = -16t^2 + 200t

To solve this quadratic equation, we can rearrange it into the standard form:
16t^2 - 200t + 380 = 0

Now we can use the quadratic formula to find the roots (t-values) of the equation: t = (-b ± √(b^2 - 4ac)) / (2a)

Here, a = 16, b = -200, and c = 380.

Plugging these values into the quadratic formula and simplifying, we get:
t = (-(-200) ± √((-200)^2 - 4 * 16 * 380)) / (2 * 16)
t = (200 ± √(40000 - 24320)) / 32
t = (200 ± √15680) / 32

Now, calculating the square root of 15680, we get:
t = (200 ± 125.25) / 32

Using both the positive and negative square root values, we have two solutions:
t1 = (200 + 125.25) / 32 = 325.25 / 32 ≈ 10.164 seconds
t2 = (200 - 125.25) / 32 = 74.75 / 32 ≈ 2.336 seconds

Rounding these values to the nearest tenth of a second, we have:
t1 ≈ 10.2 seconds
t2 ≈ 2.3 seconds

Therefore, the rocket will be 380 feet above the ground approximately 2.3 and 10.2 seconds after the launch.

Answer 4

just solve

-16t^2 + 200t = 380
-16t^2 + 200t - 380 = 0
Now just use the quadratic formula to find the roots.