Question #d18ed

Jun 27, 2015

The ratio of the time of flight for the two balls will be equal to 1.

Explanation:

This is a great example of a trick question.

When you launch an object horizontally from a height $h$, the only parameter that influences that object's time of flight is the actual height, i.e. the value of $h$.

The launch velocities have no impact on the total time of flight, they only affect the horizontal distance covered by the object.

The movement of the two balls has a horizontal component and a vertical component. The same can be said for the initial velocity.

$\left\{\begin{matrix}{v}_{0 x} = {v}_{0} \cdot \cos \left(\theta\right) \\ {v}_{0 y} = {v}_{0} \cdot \sin \left(\theta\right)\end{matrix}\right.$

When you launch a ball horizontally, the angle it makes with the horizontal is equal to ${0}^{\circ}$, which implies that

$\left\{\begin{matrix}{v}_{0 x} = {v}_{0} \cdot \cos \left({0}^{\circ}\right) = {v}_{0} \\ {v}_{0 y} = {v}_{0} \cdot \sin \left({0}^{\circ}\right) = 0\end{matrix}\right.$

Since the initial velocity has no vertical component, the vertical displacement will be equal to

$- h = {\underbrace{{v}_{0 y}}}_{\textcolor{b l u e}{\text{=0}}} \cdot t - \frac{1}{2} g {t}^{2}$

$h = \frac{1}{2} g {t}^{2}$, where

$h$ - the height of the tower;
$t$ - the total time of flight.

This means that you have

$h = \frac{1}{2} g \cdot {t}_{1}^{2}$ $\to$ for the first ball;

$h = \frac{1}{2} g \cdot {t}_{2}^{2}$ $\to$ for the second ball;

Therefore,

$\cancel{\frac{1}{2} \cdot g} \cdot {t}_{1}^{2} = \cancel{\frac{1}{2} \cdot g} \cdot {t}_{2}^{2}$

${t}_{1}^{2} = {t}_{2}^{2} \implies {t}_{1} / {t}_{2} = \sqrt{1} = 1$