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Question 1172455: Student A, who is standing in the fourth floor drops the ball from the ledge of the 12-meter building. When the ball is 3 meters above the ground, student B, standing 1.8 meters tall, looks up and notices that the ball is directly above him. How long does student B have to get out of the way?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
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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Answer: approximately 0.087 seconds
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