Question

A gun is fired with muzzle velocity 1114 feet per second at a target 4950 feet away. Find the minimum angle of elevation necessary to hit the target. Assume the initial height of the bullet is 0 feet, neglect air resistance, and give your answer in degrees.

Answer 1

The elevation angle is
`\theta = 3.67^o`

Step-by-step explanation:

From the question we are told that

The velocity is
`u = 1114\ ft/s`

The range of the bullet projection is
`R = 4950 \ ft`

The initial height of bullet is
`h_1 = 0 \ ft`

Range of this bullet projection is mathematically represented as

`R = (u^2 sin (2 \theta ))/(g)`

=>
`\theta = (sin ^(-1) [(R * g )/( u^2) ])/(2)`

Here
`g = 32 ft/s^2`

substituting values

`\theta = (sin ^(-1) [(4950 * 32 )/( 1114^2) ])/(2)`

`\theta = 3.67^o`