anonymous
  • anonymous
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?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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John_ES
  • John_ES
(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, \[\alpha=\frac{1}{2}\arctan\left(g\frac{x}{250^2}\right)=\frac{1}{2}\arctan\left(9.8\frac{120}{250^2}\right)=9.4\cdot10^{-3}\ rads\]

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