We know that the horizontal velocity is given by $v_x = v \cos(\theta)$, where $v$ is the initial velocity and $\theta$ is the launch angle. In this case, $v = 20 \, \mathrm{m/s}$ and $\theta = 45^{\circ}$, so $v_x = 20 \cos(45^{\circ})$. The distance to the wall
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