physics
posted by Emily .
A small steel ball bearing with a mass of 28.0 g is on a short compressed spring. When aimed vertically and suddenly released, the spring sends the bearing to a height of 1.25 m. Calculate the horizontal distance the ball would travel if the same spring were aimed 34.0 deg from the horizontal.

energy in spring: m*g*h
velocity of the ball= sqrt 2gh check that.we will use this later.
now calculate the time in air launched at 34deg
hf=hi+Vi*sin34*time4.9time^2
solve for time in air, t.
horizontal distance= Vi*cos34*t 
my answer for this question came out to be 1.16 m. However when I plug it into the Physics homework program it say incorrect.
My equation for t was:
1.25 m = 0m + 4.952sin34t  4.9t^2 and then I did the quadratic equation and got t= .28258
I then did the horizontal distance equation using 4.952 as my Vi.
What am I doing wrong where I still don't get the correct answer? 
The way to solve this is to use potential and kinetic energy and the formula R = (v^2/g)sin2(theta).
mgh=(1/2)mv^2
v=sqrt(2gh)
R=(sqrt(2gh)^2/9.8)sin2(theta)
= ((4.95)^2/9.8)sin(2*34)
=3.27m
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