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The additional time pass before the ball passes the tree branch on the way back down is 1.913 .
Consider the displacement equation
s(t) = (-g/2)t^2 + v(0)t + s(0)
where g = 9.8 m/s/s is the acceleration due to gravity,
v(0) = 15 m/s is the initial velocity, and
s(0) = 0 is the staring position (the ground).
We want to know when s(t) = 7 m. Thus,
7 = (-9.8/2)t^2 + 15t
We solve for t by solving the quadratic
(-9.8/2)t^2 + 15t - 7 = 0
Using the Quadratic Formula, we get the two roots
t1 = .574 seconds
t2 = 2.487 seconds
These two values represent the times when the ball passes the tree branch. Therefore, the total time between passing the branch is the difference between 2.49 seconds and 0.57 seconds.
Thus, the ball is at 7 meters at both times. The additional time after the first instance is
d = t2 - t1 = 2.487 - .574 = 1.913 seconds
To know more about " Displacement Equation"
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