# Does it take less time to drop a ball than fire one horizontally (with $90^{\circ}$) [closed]

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 $v^2 = 2gH$

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

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.

## closed as off-topic by ACuriousMind♦, Martin, Bernhard, Ryan Unger, Sebastian RieseSep 13 '15 at 19:22

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