9514 1404 393
Answer:
16 seconds
Step-by-step explanation:
You want to find t such that h(t) = 0
h(t) = -16t^2 +4096 . . . equation for height
0 = -16t^2 +4096 . . . . .equation for height = 0
16t^2 = 4096 . . . . add 16t^2
t^2 = 256 . . . . . . . divide by 16
t = 16 . . . . . . . . . . . take the positive square root
It will take 16 seconds for the camera to hit the ground.
I read this as -5e⁻⁴ˣ⁺²+3=½log₁₀(x²+1).
If this is not right the steps below can guide you to a possible solution when the correct formula is used.
My calculator is a Casio fx-85GT PLUS which has a function facility (MODE 3: TABLE).
I input the function: -5e⁻⁴ˣ⁺²+3-½log₁₀(x²+1). I assumed that log was log to base 10 rather than natural log ln.
The facility asks me to input a start value, end value and step value. I chose -10, 10 and 1 initially.
After a few seconds of evaluating a table of results the calculator showed me the results of plugging in x=-10 up to 10 in steps of 1.
The results showed the evaluation for each of the 21 values and I looked for the sign of the result to change.
For x=0 the result was -33.95 approx and for x=1 it was 2.1728 approx. So between 0 and 1 there is a zero, a solution for x.
I returned to the function and changed the parameters to start=0, end=1, step=0.1 to begin the next evaluation.
This time the sign change occurred between x=0.6 and 0.7.
I returned to the function with parameters: start=0.6, end=0.7, step=0.01.
The sign change occurred between 0.63 and 0.64, the start and end parameters with step=0.001 for the next iteration.
The solution is between 0.633 and 0.634.
Repeat the process one more time with step=0.0001. The sign change was between 0.6338 and 0.6339 so the solution to three dec places is 0.634.
