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

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

12 Online Users


New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.