# A bullet leaves the muzzle of a gun at 250 m/s to hit a target 120m away at the level of the muzzle (1.9 m above ground), the gun must be aimed above the target. (Ignore air resistance.) How far above the target is that point?

Jun 12, 2015

The distance is $1.11 m$.

#### Explanation:

To answer at the question we have to find the angle $\alpha$ above the horizontal, at which we have to shoot.

The motion is a parabolic motion, that is the composition of two motion:

the first, horizontal, is an uniform motion with law:

$x = {x}_{0} + {v}_{0 x} t$

and the second is a decelerated motion with law:

$y = {y}_{0} + {v}_{0 y} t + \frac{1}{2} g {t}^{2}$,

where:

• $\left(x , y\right)$ is the position at the time $t$;
• $\left({x}_{0} , {y}_{0}\right)$ is the initial position;
• $\left({v}_{0 x} , {v}_{0 y}\right)$ are the components of the initial velocity, that are, for the trigonometry laws:
${v}_{0 x} = {v}_{0} \cos \alpha$
${v}_{0 y} = {v}_{0} \sin \alpha$
($\alpha$ is the angle that the vector velocity forms with the horizontal);
• $t$ is time;
• $g$ is gravity acceleration.

To obtain the equation of the motion, a parabola, we have to solve the system between the two equation written above.

$x = {x}_{0} + {v}_{0 x} t$
$y = {y}_{0} + {v}_{0 y} t + \frac{1}{2} g {t}^{2}$.

Let's find $t$ from the first equation and let's substitue in the second:

$t = \frac{x - {x}_{0}}{v} _ \left(0 x\right)$

$y = {y}_{0} + {v}_{0 y} \frac{x - {x}_{0}}{v} _ \left(0 x\right) - \frac{1}{2} g \cdot {\left(x - {x}_{0}\right)}^{2} / {v}_{0 x}^{2}$ or:

$y = {y}_{0} + {v}_{0} \sin \alpha \frac{x - {x}_{0}}{{v}_{0} \cos \alpha} - \frac{1}{2} g \cdot {\left(x - {x}_{0}\right)}^{2} / \left({v}_{0}^{2} {\cos}^{2} \alpha\right)$ or

$y = {y}_{0} + \sin \alpha \frac{x - {x}_{0}}{\cos} \alpha - \frac{1}{2} g \cdot {\left(x - {x}_{0}\right)}^{2} / \left({v}_{0}^{2} {\cos}^{2} \alpha\right)$

To find the range we can assume:

$\left({x}_{0} , {y}_{0}\right)$ is the origin $\left(0 , 0\right)$, and the point in which it falls has coordinates: $\left(0 , x\right)$ ($x$ is the range!), so:

$0 = 0 + \sin \alpha \cdot \frac{x - 0}{\cos} \alpha - \frac{1}{2} g {\left(x - 0\right)}^{2} / \left({v}_{0}^{2} {\cos}^{2} \alpha\right) \Rightarrow$

$x \cdot \sin \frac{\alpha}{\cos} \alpha - \frac{g}{2 {v}_{0}^{2} {\cos}^{2} \alpha} {x}^{2} = 0 \Rightarrow$

$x \left(\sin \frac{\alpha}{\cos} \alpha - \frac{g}{2 {v}_{0}^{2} {\cos}^{2} \alpha} x\right) = 0$

$x = 0$ is one solution (the initial point!)

$x = \frac{2 \sin \alpha \cos \alpha {v}_{0}^{2}}{g} = \frac{{v}_{0}^{2} \sin 2 \alpha}{g}$

(using the double-angle formula of sinus).

Now let's find, finally $\alpha$.

sin2alpha=(gx)/v_0^2=(9.8*120)/250^2=0.0188°rArr

2alpha_1=arcsin0.0188rArr2alpha=1,078°rArralpha=0.53° and

2alpha_2=180°-arcsin0.0188=178.92°rArralpha=89.47°.

The solution are both valid, but the most reasonable is the first.

Using trigonometry, now, we can find the answer.
There is a right-angled triangle with an acute angle of 0.53°, and a cathetus of $120 m$. The other cathetus is

$c = 120 \cdot \tan 0.53 = 1.11 m$.