Question

The height of a rocket, h, is increasing at a constant rate of 18 feet per second.

1. If its height at 5 seconds is 118 feet, then write an equation for h as a function of time, t, in seconds since it was fired.

2. How many seconds will it take for the rocket to be 190 feed above ground?

3. How many feet above the ground was the rocket when it was launched?

PLEASE SHOW WORK AND EXPLAIN, THANKS!


RANDOM ANSWERS WILL BE MODERATED!!!

Answer 1

Explanation:

I'd use calculus to address this problem. Not knowing whether or not you're in calculus, I'm introducing the standard equation for vertical motion subject to gravity:

h(t) = h(0) + v(0)*t - 9.8 t^2

If its height at 5 seconds is 118 feet, then write an equation for h as a function of time, t, in seconds since it was fired. That would be:

h(5) = h(0) + (18 ft/sec)(5 sec) - (1/2)*(32 ft/sec^2)(5 sec)^2 = 118 ft

Our job here is to determine the initial height, h(0). Performing the indicated mult., we get:

h(0) + 90 ft - 400 ft = 118 ft

This results in h(0) = 118 ft - 90 ft + 400 ft, or 428 ft.

Question 1:

The desired equation is h(t) = 428 ft + (18 ft/sec)*t - (16 ft/sec^2)*t^2.

Question 2:

Assuming that this equation is correct, we can set it = to 190 ft and solve for the time, t:

190 ft = 428 ft + (18 ft/sec)*t - (16 ft/sec^2)*t^2. We need to solve this quadratic equation for t:

-16t^2 + 18t + 428 - 190 = 0, or -16t^2 + 18t + 238 = 0

Determining the coefficients of the t powers: a = -16, b = 18 and c = 238

Then the determinant is b^2-4(a)(c), which here is:

18^2-4(-16)(238) = 15556, and the square root of that is 124.7.

Thus, the two possible times are:

-18 plus or minus 124.7 -18 plus or minus 124.7

t = --------------------------------------- = ---------------------------------

2(-16) -32

Thus, t = -3.34 sec and t = 4.46 sec

The rocket will be 190 ft above the ground after 4.46 sec.

Question 3: Just take that '428 ft' from the equation found in Question 2: 428 ft. above the ground when launched

Answer 2

1) Equation for height, h(t) = (18t + 28) ft at time t seconds.

2)After 9 seconds height of rocket is 190 ft

3) Initial height of rocket is 28 ft

Explanation:

Rate of increasing of height = 18 ft/s

1) Given that at t = 5 seconds height is 118 feet,

That is

h(5) = 118 ft

We know that height of rocket is increasing 18 ft in each second, and it will also have an initial height.

So

h(t) = 18t + h₀

118 = 18 x 5 + h₀

h₀ = 118 -90 = 28 ft

Equation for height, h(t) = (18t + 28) ft at time t seconds.

2) Now we need to find time when height of rocket is 190 ft

h(t) = (18t + 28) = 190

18t = 190-28 = 162

t = 9 seconds

After 9 seconds height of rocket is 190 ft

3) Here we need to find initial height of rocket, that is t = 0 seconds

h(0) = (18x0 + 28) = 28 ft

Initial height of rocket is 28 ft