I got you b, V(final)^2=V(initial+2acceleration*displacement So this turns to (0m/s)^2=(50m/s)^2+2(9.8)(d) so just flip it all around to isolate d so you get -(50m/s)^2/2(9.8) = d so you get roughly 12.7555 meters up
... Gravity makes it go 9.8 m/s slower each second. So it runs out of speed in (50/9.8) = 5.1 seconds after the shot.
... At the beginning of the shot, its speed is 50 m/s. When it runs out of speed, its speed is zero. Its AVERAGE speed during the whole 5.1 seconds is 25 m/s .
... Going straight up for 5.1 seconds, at an average speed of 25 m/s, it rises to an altitude of
(25 m/s) x (5.1 sec) = 127.5 meters .
(You said you have an answer of 125 meters. Whoever worked out that one probably used 10 m/s² for the acceleration of gravity, instead of the 9.8 that I used.)
Of course, once the arrow "runs out of speed", it starts falling, faster and faster, until it hits the ground. You could calculate how long it takes to fall, and how fast it's going when it hits the ground. But all of that is for another question.
