### Task

A gun is intended to hit an elevated target at a given angle. How big does the muzzle velocity v_{0} have to be? What is the minimum size of the angle for the present case? (The air resistance and the Coriolis force can be neglected.)

### Solution

A solution video will be published soon on this channel.

The accelerations in direction x and direction y are

\[ \require{cancel} \]\[ \tag{1} \ddot{x} = 0 \]

\[ \tag{2} \ddot{z} = -g \]

The integration of both accelerations leads to the velocity

\[ \require{cancel} \]\[ \tag{3} \dot{x} = c_1 \]

\[ \tag{4} \dot{z}(t) = -gt + c_2 \]

and the location.

\[ \require{cancel} \]\[ \tag{5} x(t) = c_1t + c_3 \]

\[ \tag{6} z(t) = -\frac{1}{2}gt^2 + c_2t + c_4\]

The position of the projectile at time t = 0 in the x-direction is equal to zero, from which c_{3} can be calculated.

\[ \tag{7} x(t=0) = 0 \]

\[ \tag{8} 0 = \bcancel{c_1t} + c_3 \]

\[ \tag{9} c_3 = 0 \]

The z-position at time t = 0 is also zero, which results in c_{4}.

\[ \tag{10} z(t=0) = 0 \]

\[ \tag{11} 0 = \bcancel{-\frac{1}{2}gt^2} + \bcancel{c_2t} + c_4 \]

\[ \tag{12} c_4 = 0 \]

The speed in the x direction at time t = 0 is the horizontal component of v_{0}, which delivers c_{1}.

\[ \tag{13} \dot{x}(t=0) = v_0 \cdot cos \alpha \]

\[ \tag{14} c_1 = v_0 \cdot cos \alpha \]

\[ \tag{15} \dot{x} = v_0 \cdot cos \alpha \]

The speed in the z-direction is the vertical component of v_{0}, so that c_{2} can be determined.

\[ \tag{16} \dot{z}(t=0) = v_0 \cdot sin \alpha \]

\[ \tag{17} v_0 \cdot sin \alpha = \bcancel{-gt} + c_2 \]

\[ \tag{18} c_2 = v_0 \cdot sin \alpha \]

With the integration constants determined so far, the functions for the position of the projectile now look like this:

\[ \require{cancel} \]\[ \tag{19} x(t) = v_0 \cdot cos \alpha \cdot t \]

\[ \tag{20} z(t) = -\frac{1}{2}gt^2 + v_0 \cdot sin \alpha \cdot t\]

The time of impact is referred to as T in the following. The x position at time t = T is 2a.

\[ \require{cancel} \]\[ \tag{21} x(t=T) = 2a \]

\[ \tag{22} 2a = v_0 \cdot cos \alpha \cdot T \]

\[ \tag{23} T = \frac{2a}{v_0 \cdot cos \alpha} \]

The z-position at time T is a.

\[ \require{cancel} \]\[ \tag{24} z(t=T) = a\]

\[ \tag{25} a = -\frac{1}{2}gT^2 + v_0 \cdot sin \alpha \cdot T\]

The calculated T (equation 23) is now inserted and the equation can be solved for v_{0}.

\[ \tag{26} a = -\frac{1}{2}g \cdot \left( \frac{2a}{v_0 \cdot cos \alpha} \right)^2 + v_0 \cdot sin \alpha \cdot \frac{2a}{v_0 \cdot cos \alpha}\]

\[ \tag{27} a = -\frac{1}{2}g \cdot \frac{4a^2}{v_0^2 \cdot cos^2 \alpha} + sin \alpha \cdot \frac{2a}{cos \alpha}\]

\[ \tag{28} a = -\frac{1}{2}g \cdot \frac{4a^2}{v_0^2 \cdot (cos(2\alpha)+1) } + 2a \cdot tan \alpha\]

\[ \tag{29} 4a \cdot tan \alpha -2a = g \cdot \frac{4a^2}{v_0^2 \cdot (cos(2\alpha)+1) } \]

\[ \tag{30} \frac{(4a \cdot tan \alpha -2a)\cdot (cos(2\alpha)+1)}{4a^2g} = \frac{1}{v_0^2 } \]

\[ \tag{31} v_0 = \sqrt{\frac{4a^2g}{(4a \cdot tan \alpha -2a)\cdot (cos(2\alpha)+1)}} \]

#### Determination of the minimum angle

Consider the two terms in brackets below the brackets in equation 31. If one of the two becomes zero, the equation no longer provides a solution. So it applies

\[ \require{cancel} \]\[ \tag{32} 4a \cdot tan \alpha -2a > 0 \]

\[ \tag{33} \alpha > arctan \frac{1}{2} \]

\[ \tag{34} \alpha > 26.57° \]

as

\[ \require{cancel} \]\[ \tag{35} cos(2\alpha)+1 > 0 \]

\[ \tag{36} \alpha > \frac{arccos(-1)}{2} \]

\[ \tag{37} \alpha > 180° \]

Obviously, the minimum angle is found with equation 34, while equation 37 represents the upper limit value.

#### Validation of the function

Below are two diagram plots with the x and y coordinates of the projectile for two different angles of fire (45 ° and 60 °) with the same parameters for distance and height. The destination is marked with a cross.

At a firing angle of 45 °, the projectile hits the target at the apex of its trajectory.

At a firing angle of 60 °, the projectile moves on a significantly elevated trajectory and hits the target in its downward movement.