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You are on the top of the castle wall and are dropping rocks from 18.35m. just when you release a rock, an archer located exactly below you, shoots an arrow straight up toward you with an initial velocity of 47.0m/s. The arrow hits the rock in midair. How long after you release the rock does this happen? Thanks for who ever will try. 

Guest Mar 20, 2017
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Assume height above ground where arrow hits rock is s meters.  

 

Rock:   18.35 - s = (1/2)*9.8*t2      where t = time in seconds and accn of gravity = 9.8m/sec2

 

Arrow:  s = 47*t - (1/2)*9.8*t^2

 

Replace s in the first equation using the second equation:

 

18.35 - (47t - (1/2)9.8*t^2) = (1/2)*9.8*t2

 

18.35 - 47t + (1/2)*9.8*t2 = (1/2)*9.8*t2

 

18.35 - 47t = 0

 

t = 18.35/47 ≈ 0.39 secs

.

The above uses the fact that for constant acceleration situations:

distance = initial velocity*time + (1/2)*acceleration*time2

Alan  Mar 20, 2017
edited by Alan  Mar 20, 2017

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