Answer:
t = 3.61 s
Step-by-step explanation:
First, let's gather all the data. We know that the initial height (yi) is 10 ft. The final height (y) would be 0, Now, the speed (V) is 55 ft/s
The expression to use, in this kind of exercise (or the model to follow) is:
y = -16t^2 + V*t + yi
You may wondering why we need to put V*t, this is because when you do the math and calculations, that expression becomes a distance.
Now, we have the speed, replacing data we have:
0 = -16t^2 + 55t + 10 (1)
This is a second grade, and we need to use the general formula to solve this which is:
t = -b ± √b^2 - 4ac / 2a
Where a, b and c are the coefficients of the expression (1), in this case, 16, 55 and 10.
Replacing data here we have:
t = -55 ± √(55)^2 - 4*(-16)*(10) / 2*(-16)
t = 55 ± √3025 + 640 / 32
t = 55 ± √3665 / 32
t = 55 ± 60.54 / 32 (2)
Now that we have (2), let's do the final math
t1 = 55 + 60.54 / 32 = 3.61 s
t2 = 55 - 60.54 / 32 = -0.17 s
As we can't use negative time, we should take the first value, and the correct answer will be 3.61 s.
