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

#1**+2 **

Assume height above ground where arrow hits rock is s meters.

Rock: 18.35 - s = (1/2)*9.8*t^{2} where t = time in seconds and accn of gravity = 9.8m/sec^{2}

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*t^{2}

18.35 - 47t + (1/2)*9.8*t^{2 = }(1/2)*9.8*t^{2}

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*time^{2}

Alan
Mar 20, 2017