Calculating Angle for Accurate Shooting

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SUMMARY

The discussion focuses on calculating the angle required for accurate shooting when a bullet is fired at a target 130 meters away. It establishes that a bullet traveling at 300 m/s will miss the target by 0.9 meters due to gravitational drop. The time taken for the bullet to reach the target is 0.43 seconds, and the gravitational acceleration acting on the bullet is 9.8 m/s². The angle to aim the gun can be determined using trigonometric functions, specifically the tangent function, to relate the distance fallen to the horizontal distance traveled.

PREREQUISITES
  • Understanding of basic physics concepts, particularly projectile motion
  • Familiarity with trigonometric functions, specifically tangent, sine, and cosine
  • Knowledge of gravitational acceleration (9.8 m/s²)
  • Ability to perform calculations involving distance, speed, and time
NEXT STEPS
  • Learn how to apply projectile motion equations in real-world scenarios
  • Study the derivation and application of the tangent function in angle calculations
  • Explore the effects of air resistance on projectile motion
  • Investigate advanced shooting techniques and their mathematical foundations
USEFUL FOR

This discussion is beneficial for hunters, sports shooters, physics students, and anyone interested in understanding the mechanics of projectile motion and accurate aiming techniques.

mathcrzy
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A hunter aims directly at a target (on the same level) 130 m away.

(a) If the bullet leaves the gun at a speed of 300 m/s, by how much will it miss the target?
.9 m

(b) At what angle should the gun be aimed so the target will be hit?
_____°
 
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1. How much time does the bullet take to reach the target, assuming its speed is constant?

2. What is the gravitational acceleration acting upon the bullet over this time?

3. What is the formula for distance dropped due to constant acceleration and zero initial velocity?

4. How far does the bullet fall due to the acceleration over the time of travel?

5. Calculate the tangent for the distance fallen vs. the distance to the target.
 
1. .43 seconds

2. 9.8m/s^2

3. ?

4. .9m in 130 m

5. ?
 
Y and X components of motion here are separate things, so in number 3 u treat the bullet falling as a separate part, where a=g. (You already have T, so just look for the formula that works) basically what u did to obtain number 4, just that you write your formula and variables down for Y component of motion.

5.I believe is trigonometry, SOH,CAH,TOA. The tangent of distance fallen vs the distance travelled, is visualized much clearer if you do a diagram, you know that distance travel is a straight parallel line on the X axis, and distance fallen is a distance in the Y axis, if you draw this you will get a right triangle. There you can calculate the angle that the bullet falls, using Angle = Tan Y/X (TOA). If you know that the bullet shot at 0 degrees will fall Y distance at a X degree angle, what would happen if you shoot it at a +X angle?

(draw diagrams, it makes it much easier to visualize)
 
i figured out the hypotenuse=130.0031153. now how do i figure out the angle?
 
Hints:

What does at2/2 equal?

Also, remember sin(Angle)=opposite/hypotenuse.
 

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