A mortar shell is fired with a muzzle speed of $325 \mathrm { ft } / \mathrm { sec }$. Find the angle of elevation of the if the shell strikes a target located $1500 \mathrm { ft }$ away. Round your answer to 2 decimal places.
Select the correct answer.
A) $12.22 ^ { \circ }$
B) $0.64 ^ { \circ }$
C) $13.51 ^ { \circ }$
D) $0.24 ^ { \circ }$ ortar

Question

Quizplus · Accepted Answer

We can use the range equation to solve this problem: 
$$R=\frac{v_0^2}{g}\sin(2\theta)$$
where $R$ is the range (1500 ft), $v_0$ is the muzzle speed (325 ft/sec), $g$ is the acceleration due to gravity (32.2 ft/sec$^2$), and $\theta$ is the angle of elevation.

Substituting in the given values, we can solve for $\theta$: 
$$1500=\frac{(325)^2}{32.2}\sin(2\theta)$$
$$\sin(2\theta)=\frac{1500\times32.2}{(325)^2}$$
$$\sin(2\theta)\approx0.9269$$
$$2\theta\approx64.53 ^{\circ}$$
$$\theta\approx32.27 ^{\circ}$$

However, this is the angle of elevation from the horizontal, so we need to subtract it from 90$^{\circ}$ to get the angle of elevation from the ground: 
$$\theta_{ground}=90^{\circ}-32.27^{\circ}\approx 57.73^{\circ}$$

Rounding to 2 decimal places, we get:
$$\theta_{ground}\approx 57.73^{\circ}$$

The closest answer choice is C: $13.51^{\circ}$, which is clearly incorrect.

A Mortar Shell Is Fired with a Muzzle Speed Of 325