Help w/Physics ?

  • #1
The question says that a hunter aims directly at a target (on the same level) 120 m away. If the bullet leaves the gun at a speed of 250m/s, by how much will it miss the target?

So my values are v_o=250m/s, r=120m, and angle=0 since they're on the same level, right? And I need to know v to subtract from 120m.

I would just like to know what equation to use, not the answer to the problem. Thank you.
 

Answers and Replies

  • #2
HallsofIvy
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The acceleration downward is 9.8 m/s2 so dv/dt= -9.8.
Integrating (or, since this is a constant, just mutiplying)
v= -9.8t+ initial vertical velocity= -9.8t+ 0 so v= dy/dt= -9.8t.
Integrating that, y= -4.9t2+ initial height= -4.9t2.

Neglecting air resistance (which we have to since there is no information on air resistance) there is no horizontal acceleration:
a= dv/dt= 0 so v= initial horizontal velocity= 250 m/s. and then
x= 250t. To go 120 m, requires that 250t= 120. Solve that equation for t and use that t in y= -4.9t2 to find how much the bullet has dropped.
 

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