Help would be greatly appreciated

  • Thread starter oxblackout12
  • Start date
In summary, a hunter is aiming directly at a target 150 m away and the bullet leaves the gun at a speed of 290 m/s. The bullet is deflected downward by the force of gravity, so the time it takes to reach the target and the vertical distance traveled must be calculated. Using these values, the angle of the gun can be determined using trigonometry.
  • #1
oxblackout12
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1. A hunter aims directly at a target (on the same level) 150 m away.



2. (a) If the bullet leaves the gun at a speed of 290 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?





please help i am completely lost on this
 
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  • #2
oxblackout12 said:
1. A hunter aims directly at a target (on the same level) 150 m away.

2. (a) If the bullet leaves the gun at a speed of 290 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?

please help i am completely lost on this


You should take into account that the bullet is deflected downward by the force of gravity. Calculate the time it takes the bullet to reach the target and the vertical distance traveled by the bullet during this time in free fall.

Eugene.
 
  • #3
(assuming no air resistance)

t = (d / v) for horizontal displacement. The bullet is not accelerating through the horizontal motion, only the vertical motion, so don't worry about a change in speed.

With time, you can find how much gravity would cause it to drop.
 
  • #4
Hi, sorry to gravedig, but I have a question on the second part of this problem. I figured out that the displacement is 1.31 m, but how do I go about finding the angle of the gun? I know you use sins, etc., but I can't figure out how you do that to incorperate 1.31 m without changing the velocity, and thus changing the dynamics of the question.
 
  • #5
Hi Anony-mouse,

Anony-mouse said:
Hi, sorry to gravedig, but I have a question on the second part of this problem. I figured out that the displacement is 1.31 m, but how do I go about finding the angle of the gun? I know you use sins, etc., but I can't figure out how you do that to incorperate 1.31 m without changing the velocity, and thus changing the dynamics of the question.

I don't think you incorporate the 1.31 m in the second part. You set up the problem again, and you know the horizontal displacement is 150 m and the vertical displacement is zero for the bullet's flight. You can get two equations in two unknowns (angle and time) and solve for both.
 

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