Question

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.

Answer 1

`x approx 29.9^(circ)`

Explanation:

We have: `r = frac(1)(32) (v_(i))^(2) sin(2x)`

Let's plug in the provided values of the range `r` and the initial velocity `v_(i)`:

`Rightarrow 42,240 = frac(1)(32) cdot (1250)^(2) cdot sin(2x)`

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

`Rightarrow 1,351,680 = 1,562,500 cdot sin(2x)`

`Rightarrow sin(2x) = 0.8650752`

`Rightarrow 2x = 59.89127832...^(circ)`

`Rightarrow x = 29.94563916...^(circ)`

`therefore x approx 29.9^(circ)`

Therefore, the minimum angle of elevation would be around `29.9^(circ)`.