From the top of a tower, one ball is dropped and another thrown horizontally with a velocity $v$ . How will they reach the ground?

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Assumptions and Details
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- Neglect the rotation of the Earth.

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Both situations have an initial speed, $u$ , of 0.

But let's say that the height is 10 meters

You can use the "suvat" equations to solve for both.

For the vertical ball:

$\begin{aligned} v^2 &=& u^2 + 2as \\ v &=& = \sqrt{0^2 + 2as} \\ &=& \sqrt{2 \cdot 9.81 \cdot 10} \\ &\approx& 14.00 \text{ m/s} \end{aligned}$

And to find the time taken:

$\begin{aligned} v &=& u + at \\ &=& 0 + at \\ t &=& \frac va \\ &=& 14/9.81 \\ &\approx& 1.43 \text{ s} \end{aligned}$

For the horizontal ball,

there are 2 components therefore it is a different speed

$v^2_{x}=u^2_{x} + 2as_{x}$ horizontal component.

$v^2_{y}=u^2_{y} + 2as_{y}$ vertical component.