Verified answer

Initial vertical velocity = 49 sin 30.0 m/s

Initial horizontal velocity = 49 cos 30.0 m/s [and ignoring air resistance it's constant]

a = g = 9.8 m/s²

(a) Consider the vertical motion and take UP as the positive direction:

at the max height v(f) = 0 m/s

v(f)² = v(i)² + 2as

s = [v(f)² - v(i)²] / 2a

s = [0 - (49 sin 30.0)²] / [2 * (-9.8)]

s = 30.6 m

(b) Consider vertical motion to get the time of flight ... UP as positive direction:

s = v(i)t + (1/2)at²

assuming the arrows lands at the same level as it was when shot:

0 = 49 sin 30.0 t - 4.9t²

4.9t² - 49 sin 30.0 t = 0

t(4.9t - 49 sin 30.0) = 0

t = 0 [that's when it's shot] or 4.9t = 49 sin 30.0

t = 5 s

Horizontal range = v(horizontal) x time

so horizontal distance = 49 cos 30.0 x 5 = 212 m