If a projectile (such as a bullet) is fired into the air with an initial velocity $v$ at an angle of elevation $\theta$, then the height $h$ of the projectile at time $t$ is given by
\[
h=-16 t^{2}+v t \sin (\theta)
\]
Find the angle of elevation $\theta$ of a rifle barrel, if a bullet fired at 1550 feet per second takes 4 seconds to reach a height of 1000 feet. Give your answer to the nearest tenth of a degree.
Answer
\(\boxed{\theta \approx 11.7}\) degrees is the angle of elevation of the rifle barrel
Step 1 ：Substitute the given values into the equation for the height of the projectile: \(1000 = -16(4)^2 + 1550(4)\sin(\theta)\)
Step 2 ：Solve the equation: \(1000 = -256 + 6200\sin(\theta)\)
Step 3 ：Rearrange the equation to solve for \(\sin(\theta)\): \(\sin(\theta) = (1000 + 256) / 6200 = 0.2029\)
Step 4 ：Find the angle \(\theta\) by taking the inverse sine of 0.2029: \(\theta = \sin^{-1}(0.2029)\)
Step 5 ：Using a calculator, we find that \(\theta\) is approximately 11.7 degrees
Step 6 ：\(\boxed{\theta \approx 11.7}\) degrees is the angle of elevation of the rifle barrel
