# If one bullet is fired at 30^@ and another at 60^@ with respect to the horizontal, given the appropriate amount of time for them both to travel as far as they can and hit the ground, what are their respective ranges?

Aug 6, 2015

I tried using "real" values:

#### Explanation:

Have a look:

So
${h}_{2} = 3 {h}_{1}$
${D}_{1} = {D}_{2}$

Aug 6, 2015

These are parabolic motions, and we want ${y}_{f 1}$ and ${y}_{f 2}$ when they are maximized, i.e. $t = {t}_{\text{1/2}}$. Let ${y}_{i 1} = {y}_{i 2} = 0$:

${y}_{f 1} = - \frac{1}{2} g {t}_{1}^{2} + {v}_{y 1} {t}_{1} = - \frac{1}{2} g {t}_{1}^{2} + v \sin \left({30}^{o}\right) {t}_{1}$
${y}_{f 2} = - \frac{1}{2} g {t}_{2}^{2} + {v}_{y 2} {t}_{2} = - \frac{1}{2} g {t}_{2}^{2} + v \sin \left({60}^{o}\right) {t}_{2}$

When the final height is $0$, just divide the time by two to get the time of max height. So for now:

$0 = - \frac{1}{2} g {t}_{1}^{2} + v \sin \left({30}^{o}\right) {t}_{1}$
$0 = - \frac{1}{2} g {t}_{2}^{2} + v \sin \left({60}^{o}\right) {t}_{2}$

With generic, positive $g$ and $v$ values, using the quadratic formula at ${30}^{o}$:

${t}_{1} = \frac{- v \pm \sqrt{{v}^{2} - 4 \left(- 0.5\right) \left(0\right)}}{2 \left(- 0.5\right)}$

$= \frac{- v \pm v}{2 \left(- 0.5\right)}$

$0 \sec$ is trivial. The ${30}^{o}$ full-distance time is:

$= \frac{- v - v}{- 1}$

$= \textcolor{g r e e n}{v \text{ s") => color(darkgreen)(t_"1/2" = v/2 "s}}$.

At this time, ${y}_{\text{max}}$ is:

$\textcolor{\mathrm{da} r k red}{{y}_{f 1}} = - \frac{g}{2} {\left(\frac{v}{2}\right)}^{2} + \left(\frac{v}{2}\right) \left(\frac{v}{2}\right) \textcolor{\mathrm{da} r k red}{= \left(- \frac{g}{2} + 1\right) {\left(\frac{v}{2}\right)}^{2}}$

Compare to ${60}^{o}$:

${t}_{2} = \frac{- \frac{v \sqrt{3}}{2} \pm \sqrt{{\left(v \frac{\sqrt{3}}{2}\right)}^{2} - 4 \left(- 0.5\right) \left(0\right)}}{2 \left(- 0.5\right)}$

$= \frac{- v \frac{\sqrt{3}}{2} \pm v \frac{\sqrt{3}}{2}}{2 \left(- 0.5\right)}$

$0 \sec$ is trivial. The ${60}^{o}$ full-distance time is:

$= \frac{- \frac{v \sqrt{3}}{2} - \frac{v \sqrt{3}}{2}}{- 1}$

$= \textcolor{g r e e n}{v \sqrt{3} \text{s") => color(darkgreen)(t_"1/2" = vsqrt3/2 "s}}$.

At this time, ${y}_{\text{max}}$ is:

$\textcolor{\mathrm{da} r k red}{{y}_{f 2}} = - \frac{g}{2} {\left(v \frac{\sqrt{3}}{2}\right)}^{2} + \left(v \frac{\sqrt{3}}{2}\right) \left(v \frac{\sqrt{3}}{2}\right) \textcolor{\mathrm{da} r k red}{= \left(- \frac{g}{2} + 1\right) {\left(v \frac{\sqrt{3}}{2}\right)}^{2}}$

$\textcolor{\mathrm{da} r k red}{{\overbrace{{\left(\cancel{v} \frac{\sqrt{3}}{\cancel{2}}\right)}^{2}}}^{{y}_{\text{max",2)) / underbrace((cancel(v/2))^2)_(y_("max} , 1}} = 3}$

Thus, $\textcolor{b l u e}{{y}_{\text{max",2) = 3y_("max} , 1}}$

For the horizontal distance, let ${x}_{i} = 0$:

${x}_{f} = {\cancel{- \frac{1}{2} {a}_{x} {t}^{2}}}^{0} + {v}_{x} t$

(there is no acceleration due to the gun after the bullet leaves the gun)

${x}_{f 1} = {v}_{x 1} {t}_{1} = v \cos \left({30}^{o}\right) {t}_{1}$
${x}_{f 2} = {v}_{x 2} {t}_{2} = v \cos \left({60}^{o}\right) {t}_{2}$

Using the results from earlier (the full time):

${x}_{f 1} = \left(v\right) \left(\frac{\sqrt{3}}{2}\right) \left(v\right) = \frac{\sqrt{3}}{2} {v}^{2} \text{m}$
${x}_{f 2} = \left(v\right) \left(\frac{1}{2}\right) \left(v \sqrt{3}\right) = \frac{\sqrt{3}}{2} {v}^{2} \text{m}$

Thus, $\textcolor{b l u e}{{x}_{f 1} = {x}_{f 2}}$

Aug 12, 2015

I'll use this equation of motion:

${v}^{2} = {u}^{2} + 2 a s$

At the zenith ${v}^{2} = 0$ so this becomes:

$0 = {u}^{2} - 2 g h$

For the projectile launched at ${60}^{0}$:

$0 = {\left(v \cos 30\right)}^{2} - 2 g {h}_{60}$

$2 g {h}_{60} = {v}^{2} {\left(\cos 30\right)}^{2} \text{ } \textcolor{red}{\left(1\right)}$

For the projectile launched at ${30}^{0}$:

$0 = {\left(v \cos 60\right)}^{2} - 2 g {h}_{30}$

$2 g {h}_{30} = {v}^{2} {\left(\cos 60\right)}^{2} \text{ } \textcolor{red}{\left(2\right)}$

Divide $\textcolor{red}{\left(1\right)}$ by $\textcolor{red}{\left(2\right)} \Rightarrow$

${h}_{60} / {h}_{30} = \frac{{\left(\cos 30\right)}^{2}}{{\left(\cos 60\right)}^{2}} = 2.95$

To get the range we can use:

${S}_{60} = v \cos 60 \times {t}_{60} \text{ } \textcolor{red}{\left(3\right)}$

${S}_{30} = v \cos 30 \times {t}_{30} \text{ } \textcolor{red}{\left(4\right)}$

The times of flight are different but we can get their relationship:

${h}_{60} = \frac{1}{2} \text{g"t_60^2" } \textcolor{red}{\left(5\right)}$

${h}_{30} = \frac{1}{2} \text{g"t_30^2" } \textcolor{red}{\left(6\right)}$

Divide $\textcolor{red}{\left(5\right)}$ by $\textcolor{red}{\left(6\right)} \Rightarrow$

${h}_{60} / {h}_{30} = {\left({t}_{60} / {t}_{30}\right)}^{2} = 3$

So ${t}_{60} = {t}_{30} \sqrt{3}$

Now substitute this into $\textcolor{red}{\left(3\right)}$ and divide by $\textcolor{red}{\left(4\right)} \Rightarrow$

$\frac{{S}_{60}}{{S}_{30}} = \frac{v \cos 60 {t}_{30} \sqrt{3}}{v \cos 30 {t}_{30}}$

${S}_{60} / {S}_{30} = \frac{\cos 60 \sqrt{3}}{\cos 30} = \frac{0.5 \times 1.732}{0.86}$

${S}_{60} / {S}_{30} = 1$

If you check the graphic you will see you get the same range whenever you use a complementary pair of launch angles eg 70/20 etc.