## anonymous one year ago Picture yourself in the castle of Helm's Deep from the Lord of the Rings. You are on top of the castle wall and are dropping rocks on assorted monsters that are 19.10 m below you. Just when you release a rock, an archer located exactly below you shoots an arrow straight up toward you with an initial velocity of 45.0 m/s. The arrow hits the rock in midair. How long after you release the rock does this happen?

1. anonymous

the rights to lotr are owned by Middle-Earth Enterprises. I would suggest you remove this post before legal actions are taken. Anyway, the initial velocity of the rock is 0, the acceleration of the rock is -9.8 m/s^2, the total displacement is -19.10 m. delta x = vot + at^2 -19.10 = 0 + (-9.8)(t)^2 1.948 = t^2 t for rock =1.395 s the initial velocity of the arrow is 45 m/s, the total displacement is 19.10m, and the acceleration is -9.8m/s^2 delta x = vot + at^2 19.10 = (45)(t) + at^2 19.10-45t=at^2 19.10-45=at^2 -25.9=(-9.8)(t)^2 2.642=t^2 t for arrow = 1.625 s t for arrow - t for rock = ammount of time before they meet 1.625 s - 1.395 s = 0.23 s I hope this is helpful, I may have done the final step wrong so feel free to correct me.

2. IrishBoy123

this is a bit of a trick question the fact that the arrow is decelerating is offset by the fact its target, the rick, is accelerating at the same opposite rate. but you can slog it out just to see that if call the time at which the meet T, then we use the equations of motion, especially $$x = x_o + ut+\frac{1}{2}at^2$$. for the rock we have $$x_r =19.1 - \frac{1}{2}gT^2$$ in time T for the arrow $$x_a = 0 + 45T - \frac{1}{2}g T^2$$ in that time and we know that $$x_a = x_r$$ when they meet so $$19.1 - \frac{1}{2}gT^2 = 45T - \frac{1}{2}g T^2$$