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 $D_h/\sqrt {2}$.

  • $\begingroup$ what about the comment below with regard to "other factors" remaining unclear? Is this assuming max height stays the same? Still a bit unclear. $\endgroup$ – stackseverywhere May 3 '14 at 13:51
  • $\begingroup$ Other factors: vertical distance, horizontal velocity, anything else is irrelevant. $\endgroup$ – LDC3 May 3 '14 at 13:54
  • $\begingroup$ I think I got it now. Thanks. So we can't use vy=voy+at because voy=0 and we have no "vy" to solve for as we would in a traditional projectile problem with some component "vy"? if this were a traditional kinematics problem with some initial component vy, then the answer would be that the horizontal distance is halved, correct? $\endgroup$ – stackseverywhere May 3 '14 at 14:13
  • $\begingroup$ @stackseverywhere Yes, a vertical velocity would certainly make it more complicated. $\endgroup$ – LDC3 May 3 '14 at 14:28

'Other factors' remaining the same are unclear. Note that you cannot keep both the max height and the initial velocity same. If the first is the case your solution is good, if the second your friend's.

  • $\begingroup$ I would add that the first is likely, since it is usually the initial condition unchanged, in a typical what if problem. $\endgroup$ – j824h May 3 '14 at 13:40
  • $\begingroup$ The initial velocity is horizontal. It can be unaffected. $\endgroup$ – LDC3 May 3 '14 at 13:53
  • $\begingroup$ @LDC3 I've just noticed that the problem was dealing with a horizontal projectile. I misunderstood. Thanks! $\endgroup$ – j824h May 3 '14 at 13:55

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