91Ó°ÊÓ

The given trajectory is $y=cx-\left(1+c^2\right)\left(\frac{g}{2}\right){\left(\frac{x}{v}\right)}^2.$

The velocity is $897m/sec.$ We have to find the permitted values of cin the trajectory equation if the round must clear a hill 200meters high at a distance of 2000 meters in front of the howitzer.

To find the values substitute all the values in the trajectory equation.

We will use the formula of the quadratic equation to find the value of c. By proceeding with the calculation further we get,

$$\begin{gathered} 2000c-24.37c^2-224.37=0 \\ {b}^2-4ac={\left(2000\right)}^2-4\left(-24.37\right)\left(-224.37\right) \\ {b}^2-4ac=4000000-21871.58 \\ {b}^2-4ac=3978128.42 \end{gathered}$$

Let's use the formula

$$\begin{gathered} c=\frac{-bÂ\pm \sqrt{b^2-4ac}}{2a} \\ c=\frac{-2000Â\pm \sqrt{3978128.42}}{2\left(-24.37\right)} \\ c=\frac{-2000Â\pm 1994.52}{-48.74} \\ c=\frac{-2000+1994.52}{-48.74}or\frac{-2000-1994.52}{-48.74} \\ c=0.112or81.9 \end{gathered}$$

To find the values substitute all the values in the trajectory equation.

First, convert 75kilometers to meters so, $75000m.$

$$\begin{gathered} y=cx-\left(1+c^2\right)\left(\frac{g}{2}\right){\left(\frac{x}{v}\right)}^2 \\ 200=c\left(75000\right)-\left(1+c^2\right)\left(\frac{9.81}{2}\right){\left(\frac{75000}{897}\right)}^2 \\ 200=c\left(75000\right)-\left(1+c^2\right)\left(4.905\right)\left(6990.97\right) \\ 200=75000c-\left(34290.71\right)\left(1+c^2\right) \\ 75000c-34290.71-34290.71c^2-200=0 \\ 75000c-34290.71c^2-34490.71=0 \end{gathered}$$

We will use the formula of the quadratic equation to find the value of c. By proceeding with the calculation further we get,

$$\begin{gathered} 75000c-34290.71c^2-34490.71=0 \\ {b}^2-4ac={\left(75000\right)}^2-4\left(-34290.71\right)\left(-34290.71\right) \\ {b}^2-4ac=5625000000-4703411169 \\ {b}^2-4ac=921588831 \end{gathered}$$

Let's use the formula

localid="1647614811145" $$\begin{gathered} c=\frac{-bÂ\pm \sqrt{b^2-4ac}}{2a} \\ c=\frac{-75000Â\pm \sqrt{921588831}}{2\left(-34290.71\right)} \\ c=\frac{-75000Â\pm 30357.68}{-68581.42} \\ c=\frac{-75000+30357.68}{-68581.42}or\frac{-75000-30357.68}{-68581.42} \\ c=0.651or1.536 \end{gathered}$$

So, it is possible to hit a target75km away.