Question

A hunter aims directly at a target ( on the same level) 120.0 m away. (a) If the bullet leaves the gun at a speed of 250 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) Using kinematic equations you'll have,
\[x=x_0+v_{0x}t\\ y=y_0+v_{0y}t-\frac{1}{2}gt^2\]In this case,
\[x=250t\Rightarrow t=x/250\\ y=y_0-\frac{1}{2}g(x/250)^2\]
So the bullet will fall before it reachs its objetive. The distance that it will fall is,
\[|y-y_0|=\frac{1}{2}g(x/250)^2=\frac{1}{2}9.8(120/250)^2=1.13\ m\]

(b) In order to reach the y_0 position, the bullet must be fired with an angle. For simplicity, take y_0=0. So we need y=y_0=0,
\[x=250\cos\alpha t\Rightarrow t=\frac{x}{250\cos\alpha}\\ 0=250\sin\alpha \frac{x}{250\cos\alpha}-\frac{1}{2}g\left(\frac{x}{250\cos\alpha}\right)^2\]
So,
\[\alpha=\frac{1}{2}\arctan\left(g\frac{x}{250^2}\right)=\frac{1}{2}\arctan\left(9.8\frac{120}{250^2}\right)=9.4\cdot10^{-3}\ rads\]