Answer to Question #255502 in Physics for Monieth Dela Cruz

Question

A hunter aims directly at a target (on the same level) 75.0 m away. (a) If the bullet leaves the gun at a speed of 180 m/s, by how much will it miss the target? (b) At what angle should the gun be aimed so as to hit the target?

Answer 1

(a) If the bullet leaves the gun at a speed of 180 m/s, by how much will it miss the target? 

The time of bullet's horizontal motion

$$t=\frac{d}{v_0}=\frac{75\:\rm m}{180\:\rm m/s}=0.417\: \rm s$$

The vertical dropping of the bullet for this time

$$\Delta y=gt^2/2=\rm 9.8\: m/s^2*(0.417\:s)^2/2=0.85\: m$$

(b) At what angle should the gun be aimed so as to hit the target?

The range

$$R=\frac{v_0^2\sin2\theta}{g}$$

So, the corresponding angle

$$\theta=\frac{1}{2}\sin^{-1}\left(\frac{gR}{v_0^2}\right)$$

$$\theta=\frac{1}{2}\sin^{-1}\left(\frac{9.8*75}{180^2}\right)=0.65^{\circ}$$