Answer:
Step-by-step 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
