Cannon Ball 2 !!!

For any projectile motion of a body the maximum horizontal range will be equal for two angles of projection . They are $\theta$ and $(90 - \theta)$ .

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Proof

The horizontal range covered by a body is given by the formula $R = \dfrac{2u^2\sin \theta \cos \theta}{g}$

Case 1 : Angle of projection is $\theta$

$R = \dfrac{2u^2\sin \theta \cos \theta}{g}$

Case 2 : angle of projection is $(90 - \theta)$

$R = \dfrac{2u^2\sin (90 - \theta) \cos (90 - \theta)}{g}$

$\implies R = \dfrac{2u^2\cos \theta \sin \theta}{g} = \dfrac{2u^2\sin \theta \cos \theta}{g}$

Hence we proved that for any projectile motion of a body the maximum horizontal range will be equal for two angles of projection. They are $\theta$ and $(90 - \theta)$ .

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Coming to our question it is given that $Robert$ fires at an angle $\theta$ and $Alex$ fires it an angle $(90 - \theta)$ . So, they both will have same horizontal range covered .

$\color{#3D99F6} \therefore \text{The game will be a Tie.}$

The solution is wrong. Even if there is no wind blowing, the air resistance will influence the range. The winning shot will be fired at at angle less than 45 deg to horizontal, and the other player will most certainly lose.

We can derive the formula for the horizontal range of a projectile, using the kinematics equation $d = d_0 + v_0 t + \dfrac{1}{2} a t^2$ .

Horizontal: We have $d_{x0} = 0, v_{x0} = v_0 \cos \theta, a_x = 0$ . Then

$\begin{array}{rll} d_x = v_0 t \cos \theta & & \small (1) \\ {\color{#FFFFFF} \text{filler}} \end{array}$

Vertical: We have have $d_y = 0, d{y0} = 0, v_{y0} = v_0 \sin \theta, a_y = \text{-}g$ , where $g = 9.8 m/s^2$ . Then

$\begin{array}{rll} 0 = v_0 t \sin \theta - \dfrac{1}{2} g t^2 & & \small (2) \\ {\color{#FFFFFF} \text{filler}} \end{array}$

We then proceed as follows:

$\begin{array}{rl} 2 v_0 t \sin \theta - g t^2 &= \ \ 0 & & & & & \small \text{[ Doubling (2) ]} \\ \\ t (2 v_0 \sin \theta - g t) &= \ \ 0 \\ \\ t &= \dfrac{2 v_0 \sin \theta}{g} & & \small (3) & & & \small \text{[ Since } t \neq 0 \text{ when projectile lands ]} \\ \\ d_x &= \dfrac{2 v_0^2 \sin \theta \cos \theta}{g} & & & & & \small \text{[ Substituting (3) into (1) ]} \end{array}$

Now, to answer our question, we simply need to note that $\sin (90^\circ - \theta) = \cos \theta$ and $\cos (90^\circ - \theta) = \sin \theta$ , so clearly the horizontal ranges for both of the launch angles will be equal.

Let us find the relation of the horizontal range $D$ as a function of $\theta$ and the initial velocity $v$ after firing. The projectile has traveled $\frac 12 D$ when it reaches the highest point. That is when $v\sin \theta - gt = 0$ , $\implies t = \dfrac {v\sin \theta}g$ and $\frac 12 D = v\cos \theta t = \dfrac {v^2}g\sin \theta \cos \theta$ , $\implies D = \dfrac {v^2}g \sin (2\theta)$ . Now, if $D_{\text{Robert}} = \dfrac {v^2}g \sin (2\theta)$ then $D_{\text{Alex}} = \dfrac {v^2}g \sin (2(90^\circ -\theta)) = \dfrac {v^2}g \sin (2\theta) = D_{\text{Robert}}$ . It is a tie.