Projectile Motion?

A missile is fired with an initial velocity (vi) of 1,250 feet per second and aimed at a target 8 miles away. Determine the minimum angle of elevation (x) (in degrees) of the missile if the range (r) of the missile in feet is represented by r = (1/32)((vi)^2)(sin(2x)). Roud to only 1 decimal place.

Mar 17, 2018

$x \approx {29.9}^{\circ}$

Explanation:

We have: $r = \frac{1}{32} {\left({v}_{i}\right)}^{2} \sin \left(2 x\right)$

Let's plug in the provided values of the range $r$ and the initial velocity ${v}_{i}$:

$R i g h t a r r o w 42 , 240 = \frac{1}{32} \cdot {\left(1250\right)}^{2} \cdot \sin \left(2 x\right)$

Note: Value of range was converted from miles to feet.

$R i g h t a r r o w 1 , 351 , 680 = 1 , 562 , 500 \cdot \sin \left(2 x\right)$

$R i g h t a r r o w \sin \left(2 x\right) = 0.8650752$

$R i g h t a r r o w 2 x = 59.89127832 {\ldots}^{\circ}$

$R i g h t a r r o w x = 29.94563916 {\ldots}^{\circ}$

$\therefore x \approx {29.9}^{\circ}$

Therefore, the minimum angle of elevation would be around ${29.9}^{\circ}$.