Please help me thanks

Question

A rocket is launched from the top of an 80 foot cliff with an initial velocity of 100 feet per second. The​ height, h, of the rocket after t seconds is given by the equation h=16t^2+100t+80. How long after the rocket is launched will it be 30 feet from the​ ground?

Answer 1

h(t) = - 16t² + 100t + 80 = 30; 8t² - 50t - 25 = 0; t = (25 + √(625 + 200)/8 = 6.72 sec

Answer 2

The height is given by the quadratic. The ground is equal to h= 0, also known as the x-intercept.

We are trying to find H = 30. Presumably, the "a" coefficient before x² should be negative, or this thing will rocket to infinity!! If the formula is correct, then it will never be 30 feet from the ground, and this is a trick question!

-16x²+100t + 80 = 30. It's a quadratic, so let's set to equal 0 and find the solutions.

Algebra

-16x²+100t + 50 = 0

Quadratic formula:

(-100 +/- sqr(100²-4*-16*50))/ 2*16 = -100

((-100 +/-) 114.89)/ -32

we want a positive t, and since the denominator is negative the numerator needs to be negative as well.

(-100 - 114.89)/-32 = 6.71

Again, I solved for -16, but generally these answers are a little cleaner. (whole numbers, or easier square root)

Hope this helps!

Answer 3

Hi Arson!

For this problem, you will need to substitute h=30 into your equation and solve for t. This is because we are trying to find the time at which the rocket will be 30 feet off of the ground. There are several methods you can use to solve for h:

1. Use the quadratic formula
2. Complete the square
3. Factor
4. Graph the equation and find where h=30

Please let me know if you have any questions about any one of these solution methods. I am happy to help!