A dart is being thrown at a board that is 2.00 m away. When it hits the board it is 0.306 m below the point where it was aimed and it was thrown at 8.00 m/s. How long was the dart in the air?
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.