Artillery A projectile Fired from the point $\left(0,0\right)$ at an angle to the positive x-axis has a trajectory given by

$y\text{ = }cx-\left(1\text{ + }{c}^2\right)\left(\frac{g}{2}\right){\left(\frac{x}{v}\right)}^2$

where

$x=$ horizontal distance in meters

$y=$ height in meters

$v=$ initial muzzle velocity in meters per second (m/sec)

$g=$ acceleration due to gravity $=9.81$ meters per second squared (m/sec2)

$c>0$ is a constant determined by the angle of elevation.

A howitzer Fires an artillery round with a muzzle velocity of 897 m/sec.

(a) If the round must clear a hill 200 meters high at a distance of 2000 meters in front of the howitzer, what $c$ values are permitted in the trajectory equation?

(b) If the goal in part (a) is to hit a target on the ground 75 kilometers away, is it possible to do so? If so, for what values of $c$? If not, what is the maximum distance the round will travel?

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