Solving Physics Problems - Aiming a Gun at a Target

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Homework Help Overview

The discussion revolves around a physics problem involving projectile motion, specifically concerning a hunter aiming at a target 100 meters away. The problem includes determining the miss distance of a bullet fired horizontally and the angle required to hit the target.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster describes their setup of the problem, including the coordinate system and initial conditions. They detail their calculations for the bullet's trajectory and express uncertainty about how to approach the second part of the problem regarding the angle of aim.

Discussion Status

Participants are engaging with the problem, with one member prompting the original poster to share their progress. Another participant offers insights into the relationship between initial velocities and angles, suggesting a method to find the time of flight and the necessary angle for hitting the target.

Contextual Notes

The original poster's calculations are based on the assumption that the bullet is fired horizontally, and they are exploring the implications of this setup for both parts of the problem. There is an indication of a lack of consensus on how to proceed with the second part of the question.

lmf22
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physics problem...

A hunter aims directly at a target (on the same level) 100m away.
a) If the bullet leaves the gun at a speed of 250m/s, by how much will it miss the target?
b) At what angle should the gun be aimed so the target will be hit?

Any help would be great. Thanks.
 
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What have you done so far?
 
I set up the axis where the hunter is Y meters up and the target is also Y meters up to 100m to the right.
The X component of the velocity is 250m/s and Y component is 0.

For part a)
I used the distance formula in the Y and X components so:
Y = Yi + vi*t + 1/2*a*t^2.
X = Xi + vi*t + 1/2*a*t^2.

Y=0
Vi=0
Xi=0
a in the X component = 0

From the X component of the distance formula I solved for time =0.4s.
Then I plug in the time into the Y component of the distance formula and got 39.2m.

I'm not sure where to start for part b).
 
Note that given initial speed V0, and angle a to the horizontal, we may write:
Initial horizontal velocity: V0cos(a)
Initial vertical velocity: V0sin(a)

Find the time T from the horizontal component (that is, which yields final horizontal position equal to 100)
Determine a by requiring that at this time T, final vertical position must be zero (equal to initial level)
 

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