Answer to Question #251321 in Mechanics | Relativity for Tweety

Question

A rifle is aimed horizontally through Its bore at the center of a large target 150m away. The initial velocity of the bullet is 450 m/s. a) Where does the bullet strike the target? b) To hit the center of the target, the barrel must be an angle above the line of sight. Find the angle of elevation of the barrel. We assume frictionless system and g = -9,81 m/s^2. (Hint sin(2a) = 2sin(a)cos(a))

Answer 1

a) The distance below the center of the target is

$$h=\frac{gt^2}{2}=\frac{g(R/v)^2}{2}=0.54\text{ m}.$$

b) To hit the target, the angle above the horizontal can be calculated from the range:

$$R=\frac{v^2\sin(2\theta)}{g},\\\space\\ \theta=\frac12·\arcsin\frac{Rg}{v^2}=0.21°.$$