# Question b606a

Sep 6, 2015

Their launch velocities must be in a ratio of $\sqrt{3}$.

#### Explanation:

As you know, you can break down the trajectory of an object launched at an angle $\theta$ with the horizontal into two parts, a horizontal component and a vertical component.

This means that you can do the same for its launch velocity, ${v}_{0}$, which will have two components

${v}_{0 x} = {v}_{0} \cdot \cos \theta \to$ horizontal component

and

${v}_{0 y} = {v}_{0} \cdot \sin \theta \to$ vertical component

Now, you know that the maximum heights of the two objects must be equal. You can focus solely on the vertical component of the movement, which is influenced by the gravitational acceleration, $g$.

At maximum height, the vertical component of the object's velocity will be equal to zero. This means that you can write

${\overbrace{{v}_{\text{top on y}}^{2}}}^{\textcolor{b l u e}{= 0}} = {v}_{01 y}^{2} - 2 \cdot g \cdot {h}_{1} \to$ for object 1

and

${\overbrace{{v}_{\text{top on y}}^{2}}}^{\textcolor{b l u e}{= 0}} = {v}_{02 y}^{2} - 2 \cdot g \cdot {h}_{2} \to$ for object 2

You know that ${h}_{1} = {h}_{2} = h$. For the first object, you have

${v}_{01 y}^{2} = 2 \cdot g \cdot h$

$h = {v}_{01 y}^{2} / \left(2 \cdot g\right) = {\left[{v}_{01} \cdot \sin \left({\theta}_{1}\right)\right]}^{2} / \left(2 \cdot g\right)$

$h = \frac{{v}_{01}^{2} \cdot {\left[\sin \left({30}^{\circ}\right)\right]}^{2}}{2 \cdot g} = \frac{{v}_{01}^{2} \cdot {\left(\frac{1}{2}\right)}^{2}}{2 g} = \frac{1}{8} \cdot {v}_{01}^{2} / g$

For the second object, you have

$h = {v}_{02 y}^{2} / \left(2 \cdot g\right) = {\left[{v}_{02} \cdot \sin \left({\theta}_{2}\right)\right]}^{2} / \left(2 \cdot g\right)$

$h = \frac{{v}_{02}^{2} \cdot {\left[\sin \left({60}^{\circ}\right)\right]}^{2}}{2 \cdot g} = \frac{{v}_{01}^{2} \cdot {\left(\frac{\sqrt{3}}{2}\right)}^{2}}{2 g} = \frac{3}{8} \cdot {v}_{02}^{2} / g$

Here ${v}_{01}$ and v_(02# are the launch velocities of the first object (for which ${\theta}_{1} = {30}^{\circ}$) and for the second object(for which ${\theta}_{2} = {60}^{\circ}$), respectively.

The ratio between these two initial velocities will be

$\frac{1}{\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}}} \cdot {v}_{01}^{2} / \textcolor{red}{\cancel{\textcolor{b l a c k}{g}}} = \frac{3}{\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}}} \cdot {v}_{02}^{2} / \textcolor{red}{\cancel{\textcolor{b l a c k}{g}}}$

${v}_{01}^{2} / {v}_{02}^{2} = 3 \implies {v}_{01} / {v}_{02} = \textcolor{g r e e n}{\sqrt{3}}$

Notice that this is the ratio between the values of $\sin {\theta}_{2}$ and $\sin {\theta}_{1}$, because you have

$h = \frac{{v}_{01}^{2} \cdot {\sin}^{2} \left({\theta}_{1}\right)}{2 \cdot g} \implies {v}_{01}^{2} = \frac{2 \cdot g \cdot h}{{\sin}^{2} {\theta}_{1}}$

Similarly, you have

${v}_{02}^{2} = \frac{2 \cdot g \cdot h}{{\sin}^{2} {\theta}_{2}}$

Divide these expressions to get

${v}_{01}^{2} / {v}_{02}^{2} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2 \cdot g \cdot h}}}}{\sin} ^ 2 {\theta}_{1} \cdot {\sin}^{2} {\theta}_{2} / \textcolor{red}{\cancel{\textcolor{b l a c k}{2 \cdot g \cdot h}}} = {\sin}^{2} {\theta}_{2} / {\sin}^{2} {\theta}_{1}$

This is equivalent to

${v}_{01} / {v}_{02} = \sqrt{{\sin}^{2} {\theta}_{2} / {\sin}^{2} {\theta}_{1}} = \sin {\theta}_{2} / \left(\sin {\theta}_{1}\right) = \frac{\sqrt{3}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} \cdot \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}}{1} = \textcolor{g r e e n}{\sqrt{3}}$