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 31.8 m/s and that the second arrow is fired 1.82 s after the first. determine the initial speed of the second arrow.

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

My answer is opptionc, just check with anyone else

The correct answer is c. hope this