A gun has a muzzle speed of 90 meters per second. What angle of elevation should be used to hit an object 150 meters away? Neglect air resistance and use as the acceleration of gravity.
Answer in radians.
Thanks!
Substitute the data into equation (7) here: Projectile Motion
Now solve this equation for $\displaystyle \theta$.
Hello, jffyx!
The equations for projectile motion are: . $\displaystyle \begin{array}{ccc}x \;=\; (v_o\cos\theta)t \\ y \;=\; h_o + (v_o\sin\theta)t - 4.9t^2 \end{array}$A gun has a muzzle speed of 90 meters per second.
What angle of elevation should be used to hit an object 150 meters away?
. . where: .$\displaystyle \begin{Bmatrix}h_o &=& \text{initial height} \\ v_o &=& \text{initial speed} \\ \theta &=& \text{angle of elevation} \end{Bmatrix}$
We are given: .$\displaystyle h_o = 0,\;v_o=90$
The equations are: .$\displaystyle \begin{array}{cc}x \;=\;(90\cos\theta)t & {\color{blue}[1]} \\ y \;=\;(90\sin\theta)t - 4.9t^2 & {\color{blue}[2]} \end{array}$
Since $\displaystyle x = 150$, [1] becomes: .$\displaystyle (90\cos\theta)t \:=\:150 \quad\Rightarrow\quad t \:=\:\frac{5}{3\cos\theta} $
Substitute into $\displaystyle y = 0\!:\;\;(90\sin\theta)\left(\frac{5}{3\cos\theta}\ right) - 4.9\left(\frac{5}{3\cos\theta}\right)^2 \:=\:0 $
which simplifies to: .$\displaystyle 1350\tan\theta - 122.5\sec^2\!\theta \:=\:0$
. . $\displaystyle 1350\tan\theta - 122.5(\tan^2\!\theta + 1) \:=\:0 \quad\Rightarrow\quad 122.5\tan^2\!\theta - 1350\tan\theta + 122.5 \:=\:0 $
Quadratic Formula: .$\displaystyle \tan\theta \;=\;\frac{1350 \pm\sqrt{1,\!762,\!475}}{245} \;=\;\begin{Bmatrix}0.091500952 \\ 10.92890771 \end{Bmatrix}$
. . $\displaystyle \tan\theta \,=\,0.091500952\quad\Rightarrow\quad \theta \:\approx\:5.228^o$