To find the time it takes for the dart to reach the board, we can use the equation of motion for vertical motion:
y = y0 + v0*t - 1/2 * g * t^2
Where:
y = final vertical displacement (0.306 m)
y0 = initial vertical displacement (0, since we are measuring from the top of the board)
v0 = initial velocity (8.00 m/s)
g = acceleration due to gravity (-9.8 m/s^2, assuming downward direction)
t = time
Substituting the known values into the equation, we can rearrange it to solve for time (t):
0.306 = 0 + 8.00 * t - 1/2 * 9.8 * t^2
Rearranging the equation gives us a quadratic equation:
4.9 * t^2 - 8.00 * t + 0.306 = 0
Using the quadratic formula, we can solve for t:
t = (-b ± √(b^2 - 4ac)) / (2a)
Where:
a = 4.9
b = -8.00
c = 0.306
Plugging in the values:
t = (-(-8.00) ± √((-8.00)^2 - 4 * 4.9 * 0.306)) / (2 * 4.9)
Simplifying further:
t = (8.00 ± √(64.00 - 6.00)) / 9.8
t = (8.00 ± √58.00) / 9.8
Now we can calculate the two possible values of t:
t1 = (8.00 + √58.00) / 9.8
t2 = (8.00 - √58.00) / 9.8
Calculating these values using a calculator:
t1 ≈ 1.066 seconds
t2 ≈ 0.014 seconds
Since time cannot be negative in this context, we discard t2 as an extraneous solution.
Therefore, the dart was in the air for approximately 1.066 seconds.