What Formula Should I Use to Solve a Horizontal Projectile Problem?

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Homework Help Overview

The problem involves a horizontal projectile motion scenario where a bullet is fired towards a target at a distance of 68 m with an initial speed of 186 m/s. The original poster seeks guidance on the appropriate formula to determine how much the bullet will miss the target due to gravitational drop.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss calculating the time it takes for the bullet to reach the target and the vertical drop due to gravity. There are attempts to apply the equations of motion, with some questioning the accuracy of their calculations and the interpretation of initial velocities.

Discussion Status

The discussion is ongoing, with participants providing guidance on separating horizontal and vertical components of motion. Some participants are exploring different interpretations of initial velocities and their implications on calculations, while others are verifying their results and seeking clarification on specific values.

Contextual Notes

Participants are working under the assumption that the target is at the same vertical level as the gun, which influences the initial vertical velocity component. There is also a focus on ensuring the correct application of kinematic equations in the context of projectile motion.

xelda
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A hunter aims directly at a small target (on the same level) 68 m away. If the bullet leaves the gun at a speed of 186 m/s, by how much will it miss the target?

I just need to know what formula I should use to solve this problem?
 
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First calculate the time it takes for the bullet to traverse the given distance.
Then calculate the amount by which the bullet dropped due to gravity.
 
I understand that's how to solve the problem, but I still don't think I'm doing this correctly. To calculate the time, I took 186 m/s divided by 68 m and got 2.735 s. Then I plugged the time into the equation y=yo+vyot-1/2gt^2. So it looked like 0+186m/s(2.735s)-1/2(9.8m/s^2)(2.735s)^2. I ended up with 472 but this number seems too large? Am I doing something wrong here?

Thank you for the quick response. I appreciate your help.
 
You seem to have confused a few things.
You have to treat the horizontal and vertical components independently.
First, since there is no horizontal acceleration, the horizontal velocity is constant.
To get the time traveled, use d=vt. (d=distance, v=velocity, t=time) (so t=d/v)

Then for the vertical component, you can use [itex]y(t)=y_0+vt-1/2gt^2[/itex], but
think about what v is in this equation. It's the initial component of the velocity in the y-direction.
 
Okay, I used 9.9 m/s^2 for v and my final answer came to 2.93 m. Does this seem more accurate? I guess I was way off. :x
 
What value did you find for t? And what is the initial v in the vertical direction?
 
For t I got .3656 s. I put 0 in for yo but should v also be 0 if it's the vertical component?
 
xelda said:
For t I got .3656 s. I put 0 in for yo but should v also be 0 if it's the vertical component?
If the target is on the same level as the gun, then isn't the initial velocity horizontal?
 
Oh, okay. Now it's starting to make more sense. Since the initial velocity is horizontal, then it would be 0.
 

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