Physics Question... two arrows are shot vertically upward?

Question

Two arrows are shot vertically upward. The second arrow is shot after the first one, but while the first is still on its way up. The initial speeds are such that both arrows reach their maximum heights at the same instant, although these heights are different. Suppose that the initial speed of the first arrow is 27.8 m/s and that the second arrow is fired 1.05 s after the first. Determine the initial speed of the second arrow.

Answer 1

Ok. We are gonna do this using kinematics.



You know that the two arrows reach their maximum height at the same time, so lets find that time out.



The initial speed or the first arrow is 27.8 m/s, it reaches its maximum height when its speed is 0 m/s, and gravity is -9.8 m/s^2. Pop those numbers into Vf=Vi+a*t to get 0=27.8+-9.8*t. Solve for t and get 2.8 seconds.



Now that you know the time to reach its maximum height, you have enough information to find out the initial velocity of the second arrow. Here's what you know about it: its final velocity is 0 m/s (at the maximum height), its time to reach that is 2.8 seconds, but wait! it was fired 1.05 seconds later, so take off 1.05 seconds so that its time is 1.75 seconds, and of course gravity is still the same at -9.8 m/s^2. Plug those numbers into the kinematic equation (Vf=Vi+a*t, remember?) for 0=Vi+-9.8*1.75 and solve for Vi to get.......



17.15 m/s !!