Question 1172455
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We'll ignore air resistance as it greatly complicates the problem. 
We don't have enough info even if we wanted to account for air resistance.


We'll use the free fall equation which is
y = 0.5*g*t^2
the variables are
y = vertical distance the ball travels
g = acceleration of gravity
t = elapsed time


On earth, the acceleration of gravity is roughly 9.81 m/s^2
So g = 9.81 approximately


The free fall equation updates to
y = 0.5*g*t^2
y = 0.5*9.81*t^2
y = 4.905t^2
which is approximate.


If student B wasn't in the way, then the ball would travel 12 meters. 
If student B is there, and the ball hits them, then the ball travels 12-1.8 = 10.2 meters


Let's determine how long it takes for the ball to travel 10.2 meters


Plug in y = 10.2 and solve for t
y = 4.905t^2
10.2 = 4.905t^2
4.905t^2 = 10.2
t^2 = 10.2/4.905
t^2 = 2.07951070336391
t = sqrt(2.07951070336391)
t = 1.44205086712082
t = 1.442


It takes about 1.442 seconds for the ball to hit student B


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Now let's calculate the time it takes for the ball to travel 12-3 = 9 meters. 
This is the moment in time when student B looks up and sees the ball 3 meters above the ground.


Plug in y = 9 and solve for t
y = 4.905t^2
9 = 4.905t^2
4.905t^2 = 9
t^2 = 9/4.905
t^2 = 1.8348623853211
t = sqrt(1.8348623853211)
t = 1.35457092295719
t = 1.355


It takes about 1.355 seconds for the ball to drop 9 meters.


Subtract the time values we calculated
1.442 - 1.355 = 0.087


So the student B has approximately 0.087 seconds to react and get out of the way after they spot the ball.


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<font color=red>Answer: approximately 0.087 seconds</font>
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