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.
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