# When solving kinematics questions, how to find time in air?

I was trying to help a friend with this question:

A ball is shot horizontally off a cliff. What would happen to the horizontal distance it travels if the acceleration due to gravity were doubled and all other factors remain the same?

My initial reaction was to do

$-\frac{v_0}{g} * 2$ to get the total time in air, if $g$ is doubled, the time in air is halved, and therefore the horizontal distance is halved $(v_x \cdot *t)$ $t$ is halved while $v_x$ stays constant.

My friend was using $y=\frac{1}{2}at^2$ to solve for half $t= \sqrt{\frac{2y}{g}}$ and saying that the time would decrease by a factor of $\sqrt 2$. How do you reconcile these two formulas to find the time in air? I have traditionally used my first method and gotten these questions right, but I am having a brain delay this morning on why you can't use the 2nd formula?

Any clarifications would be appreciated!

Your friend is right. The time to travel the vertical distance is $t/\sqrt {2}$, which corresponds to the horizontal distance traveled as $D_h/\sqrt {2}$.