How to solve acceleration due to gravity without time? If you have a dot on a computer screen that you launch from 90 pixels high with a velocity of 100 at an angle of 0, and it lands on the bottom of the screen 150 pixels away, what is its acceleration due to gravity?
Recording time from launch with a stopwatch is inaccurate, so I'd rather solve this without time as it's not necessary (I know it isn't because I've solved it before but have forgotten)
Once again there should be no time variable, to solve this using time is all over the internet, I need the equation without time, as can be done when you know initial velocity (100), initial height (90 pixels), horizontal distance (150 pixels), and vertical distance that it fell (90 pixels) from the known launch angle (0)
 A: We do not need to be given the travel time, since the information supplied is enough to determine that. 
If the time to fall is $t$, then the horizontal distance travelled is $100t$. But this is $150$, so $t=1.5$. 
If the acceleration is $a$, then the distance fallen in time $t$ is $\frac{1}{2}at^2$. This is $90$.
Thus $\frac{1}{2}a(1.5)^2=90$. That gives $a=\dfrac{(2)(90)}{(1.5)^2}$.
Remark: A similar calculation can be made for any known launch angle $\theta$. We have tacitly assumed that acceleration is constant. In particular, the calculation neglects the (possibly very considerable) effect of air resistance.
Mentioning time less: Let $v_{\text{av}}$ be the average downward velocity. Then $\frac{v_{\text{av}}}{90}=\frac{100}{150}$. But $v_{\text{av}}=\frac{1}{2}a\frac{150}{100}$. It follows that $a=(2)(90)\left(\frac{100}{150}\right)^2$. 
A: Acceleration is defined as the second derivative of position with respect to time, or the first derivative of velocity with respect to time. Time is an intrinsic property of the measurement. You cannot do it without time. 
