Physics help (:An arrow is shot vertically upward and then 1.81 s later passes the top of a tree 43.2 m high

This last problem on my homework is horrible, I cannot figure it out. Please help if you can, here's the question: An arrow is shot vertically upward and then 1.81 s later passes the top of a tree 43.2 m high. How much longer will it go?

time: ?
Height; ?


Thanks again!

time = ?
initial velocity(u) = ?
acceleration = -9.8 (due to gravity)
distance(s) = 43.2
time = 1.81...

the key is to find initial velocity

s = ut + 1/2 at^2
43.2 = u(1.81) + 1/2(-9.8)(1.81^2)


re-arrange to get [43.2 - 1/2(-9.8)(1.81^2) ] /1.81 = u
when u calculate u, you wanna find the max height it reaches

note, if its going up in the air, and falls back down again, it will come to its heighest point, and its velocity will be zero, then it will accelerate towards the ground...

v^2 = u^2 + 2as.....v=0 at its max height
so 0 = u^2 + 2as
re arrange to get
(-u^2)/2(-9.8) = s

u will get a value for s. take 43.2 from that value, and u will have the correct answer for distance....

then, to find time

v=u+at
0=u+(-9.8)t
re-arrange to get, -u/-9.8 = t

take 1.81 from the value u get for t, this will give u the correct answer