Question

Problem 4.

A cannon has a muzzle speed of 550 feet per second. A shot is fired from

ground level at an angle of 20◦ and overshoots the target by 50 feet. At what

angle should the next shot be fired to hit the target (same muzzle speed)?

Neglect air resistance. Give numerical answers accurate to two decimal places.

Answer 1

Let $x$ be the distance to the target. Then, the horizontal distance travelled by the shot is $R = x + d$ where $d = 50ft$. On the other hand, the horizontal distance is given as follows (see https://courses.lumenlearning.com/boundless-physics/chapter/projectile-motion/):

$$R = \dfrac{u^2\sin(2\theta)}{g}$$

where $u = 550ft/s$ is the initial speed, $\theta = 20\degree$ is launching angle, and $g \approx 32.17ft/s^2$ is the gravitational acceleration.

Thus, obtain:

$$x + d = \dfrac{u^2\sin(2\theta)}{g}\\ x = \dfrac{u^2\sin(2\theta)}{g} - d\\ x + 50 = \dfrac{550^2\cdot \sin(2\cdot 20\degree)}{32.17}\\ x\approx$$

The angle $\theta _0$ , required to aim the target exactly can be found from the following equation:

$$x = \dfrac{u^2\sin(2\theta_0)}{g} = \dfrac{u^2\sin(2\theta)}{g} - d$$

Expressing $\theta _0$ from the last equation, obtain:

$$\sin(2\theta_0) = \sin(2\theta) - \dfrac{gd}{u^2}\\ \theta_0 = \dfrac12\arcsin\left( \sin(2\theta) - \dfrac{gd}{u^2} \right)\\ \theta_0 = \dfrac12\arcsin\left( \sin(2\cdot 20\degree) - \dfrac{32.17\cdot 50}{550^2} \right) \approx 19.80\degree$$

Answer. 19.80 degrees.