Question

An arrow is shot vertically upward and then 2.19 s later passes the top of a tree 36.9 m high. How much longer will the arrow travel upward, and how high will it go?

Answer 1

use the equation y = -9.8*t^2+v*t+h
we'll assume h (initial hight) is zero
v (initial velocity) is what we need to solve for at first

so, we plug in our givens:
36.9=-9.8*(2.19)^2+2.19*v
v=38.311 m/s
do some algebra to get
v=38.311

now we need to know the time when the velocity equals zero, so we take the derivative of that function to get
y'=-19.6*t+v
where y' is the instantaneous velocity

now plug and chug there
0=-19.6*t+38.311
t=1.9546

finally we can put this back into our original equation to get our final height
y=-9.8*(1.9546)^2+57.514*1.9546

your final answers are
y=75 m
t=1.9546 s