#### Answer

horizontal distance at 45 degrees is less than that at 42 degrees.

#### Work Step by Step

Lets first calculate the time it passes till the shot hits the ground.
We can use $-H=v_{0}t\sin \alpha -g\dfrac {t^{2}}{2}$ to calculate this time. Here, H is initial height of shot, $v_{0}$ is initial velocity and $α$ is initial angle of velocity with respect to the horizontal line. The minus sign indicates that the shot will travel in the (-) y direction with the magnitute of H distance in order to hit the ground.
Lets solve this equation:
$-2160m=15\dfrac {m}{s}\times \sin 42^{0}\times t-\dfrac {9.8\dfrac {m}{s^{2}}t^{2}}{2}$
After solving, we find $t=22.0448s$.
For the horizontal distance traveled by object :
$L=v_{0}t\cos \alpha $
So we get,
$L=15\dfrac {m}{s}\times \cos 42^{0}\times 22.0448s=245.737m$
Therefore,
$L_{\alpha =45}=234.47m$ And $L_{\alpha =42}=245.737m$
We prove that $L_{\alpha =42} > L_{\alpha =45}$ so $α=45^{0}$ doesnt maximize the horizontal distance when launching and the landing height is also different.