Answer :
Answer:
The elevation angle is [tex]\theta = 3.67^o[/tex]
Explanation:
From the question we are told that
The velocity is [tex]u = 1114\ ft/s[/tex]
The range of the bullet projection is [tex]R = 4950 \ ft[/tex]
The initial height of bullet is [tex]h_1 = 0 \ ft[/tex]
Range of this bullet projection is mathematically represented as
[tex]R = \frac{u^2 sin (2 \theta )}{g}[/tex]
=> [tex]\theta = \frac{sin ^{-1} [\frac{R * g }{ u^2} ]}{2}[/tex]
Here [tex]g = 32 ft/s^2[/tex]
substituting values
[tex]\theta = \frac{sin ^{-1} [\frac{4950 * 32 }{ 1114^2} ]}{2}[/tex]
[tex]\theta = 3.67^o[/tex]