Question

A projectile fired into the first quadrant from

the origin of a coordinate system will pass through the point

(x, y) at time t according to the relationship $cot\theta =\frac{2x}{2y+g{t}^2}$,

where $\theta$= the angle of elevation of the launcher and g =

the acceleration due to gravity = 32.2$ft/se{c}^2$

. An artilleryman is firing at an enemy bunker located 2450 feet

up the side of a hill that is 6175 feet away. He fires a round,

and exactly 2.27 seconds later he scores a direct hit.

(a) What angle of elevation did he use?

(b) If the angle of elevation is also given by $sec\theta =\frac{v_{0}t}{x}$

x , where$v_0$is the muzzle velocity of the weapon, find the muzzle

velocity of the artillery piece he used.

Short answer

(a) The angle of elevation is $22.3^{\circ }$.

(b) The muzzle velocity is$2940.20ft/sec$.

Step-by-step solution

Part (a) Step 1. Given Information

An artilleryman is firing at an enemy bunker located 2450 feet

up the side of a hill that is 6175 feet away. He fires a round,

and exactly 2.27 seconds later he scores a direct hit.

Part (a) Step 2. Calculation

$$\begin{gathered} hill=6175ft \\ enemy=2450ft \\ cot\theta =\frac{2\left(6175\right)}{2\left(2450\right)+(32.2)(2.27)} \\ =\frac{12350}{5065.92338} \\ =2.4379 \\ therefore, \\ \theta =co{t}^{-1}\left(2.4379\right) \\ =22.3^{\circ } \end{gathered}$$

Part (b) Step 1. Calculation

$$\begin{gathered} Substitutex,tand\theta intheequation \\ sec22.3=\frac{v_0(2.27)}{6175} \\ =0.0003676v_0 \\ \frac{sec22.3}{0.0003676}=\frac{0.0003659v_0}{0.0003676} \\ {v}_0=\frac{sec22.3}{0.0003676} \\ =\frac{1.0808}{0.0003676} \\ =2940.23 \end{gathered}$$