So I was arguing about this with my friend. If we take two balls and drop one from a certain height H and then fire another one with horizontally with some initial speed from the same height H, which one will fall faster. My friend said that they would fall at the same time but I made the following argument.

In Case I (let us drop ball A):

$$ v = u + gt $$ since we are dropping a ball, $u = 0$ $$ t = \frac{v}{g}$$ Now we use $v^2 = u^2 + 2gH$

Again $u^2 = 0$

So $ v^2 = 2gH $

$ v = \sqrt{2gH} $ $$ t = \frac{\sqrt{2gH}}{g} $$ $$ t = \sqrt{\frac{2H}{g}} $$

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In Case II( let us fire ball B horizontally so it travels in a trajectory):

Since only half the trajectory is being travelled, we use $\frac{T}{2}$ $$ \frac{T}{2} = \frac{usin\theta}{g}$$ $$ \frac{T}{2} = \frac{usin 90^{\circ} }{g}$$ $$ T = \frac{u}{g}$$ now H is the maximum height, we have a formula for it. $$ H = \frac{u^2sin^2\theta}{2g} $$ $$ H = \frac{u^2}{2g} $$ $$ u = \sqrt{2Hg}$$ Now substituting $u$ in $T$ $$ T = \frac{\sqrt{2Hg}}{g} $$ $$ T = 2\sqrt{\frac{2H}{g}}$$

So T = t. They both fall at the same time.


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They will hit at the same time; you can see this via Galilean relativity. If you are on a train moving at high speed and drop a ball straight down from your own point of view, it falls normally (indistinguishably from how it falls when you are standing on terra firma). From the point of view of someone outside the train, the ball starts off with a high horizontal velocity. Since you agree on how long it takes the ball to fall, it takes balls the same time to fall whether fired horizontally or not.

I find your argument unclear; it is just a long list of equations. Without some explanation of what you intend for these equations to mean, we can't critique your work.

  • $\begingroup$ Thanks I checked my equations again, I had made a few mistakes, I got time taken same for both. $\endgroup$ – Abhishek Mhatre Sep 9 '15 at 16:07
  • $\begingroup$ This was exactly the answer that I came here to post. Symmetry principles are always the best way to go! $\endgroup$ – Danu Sep 10 '15 at 16:40
  • $\begingroup$ Of course this explanation would not work when air friction is considered. $\endgroup$ – Bernhard Sep 13 '15 at 13:29

Reformulating Galilean relativity slightly: horizontal and vertical motion are completely independent of each other. The body falls vertically, whether or not it also moves horizontally.


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