Answer
13.2 m/s
Work Step by Step
Let's apply equation $ V=u+at$ to the first arrow to find the time required to reach its maximum height.
$\uparrow V=u+at$ ; Let's plug known values into this equation.
$0\space m/s=25\space m/s -9.8\space m/s^{2}\times t$
$t=\frac{(0-25)m/s}{-9.8\space m/s^{2}}=2.55\space s$
Given that the second arrow is shot 1.20 s after the first arrow. Therefore, since both arrows
reach their maximum height at the same time, the second arrow reaches its maximum height
$time=2.55\space s-1.20\space s=1.35\space s$
Let's apply equation $V=u+at$ to find the second arrow to find the initial speed,
$\uparrow V=u+at$ ; Let's plug known values into this equation.
$0\space m/s=u+(-9.8\space m/s^{2})\times1.35\space s$
$u=13.2\space m/s$
The initial speed of the second arrow = 13.2 m/s
