Angle of Bullet Fired: Solving for the Initial Angle

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Homework Help Overview

The problem involves determining the angle at which a bullet is fired from a gun, given its initial velocity of 300 m/s and a time of flight of 3 seconds, while assuming no air resistance. The context is rooted in kinematics and projectile motion.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss breaking down the velocity into x and y components and consider the effects of gravity on the y component. Questions arise regarding the initial vertical velocity and the application of kinematic equations to find the angle of projection.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and questioning assumptions about initial velocities. Some guidance has been offered regarding the use of kinematic equations, but no consensus has been reached on the correct approach or calculations.

Contextual Notes

There is a lack of explicit information regarding the initial vertical velocity, leading to uncertainty in the calculations. Participants are also navigating the implications of assuming no air resistance in their analysis.

Physics08
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Hi,

Could somebody please tell me what formula(s) to use to find the answer to the following problem?

A bullet is fired from a gun at 300 m/s. It hits the ground 3s later. At what angle (in degrees) above the horizon was the bullet fired? (Assume no air resistance)

Thank you in advance
 
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well you have velocity and a time. Think about breaking things up in the x and y components.

In this case, does the x component of velocity ever change?

The Y component of your velocity changes and it changes at a rate of 9.8m/s^2

Since its constant acceleration, use a constant acceleration equation that relates the time, starting velocity, final velocity, and acceleration
 
OK, how 'bout this?

V(y-final) = V(y-initial) + a(y)t

V(y-final) = -9.8m/s^2 * 3

V(y-final) = -29.4 m/s

then find angle using inverse tangent

Angle = tan-1 (-29.4/300)

Angle = -5.52

Is this correct?
 
Let me know if I understand the question correctly. Your gun is set on the ground and mounted at an angle that you need to find. The bullet is shot out at 300 m/s and stays in the air for 3 seconds.

If that is the case, is your initial y velocity correct?
 
here's a question to ask based on the work you just did, is the initial y-velocity zero? Do you know the initial y-velocity?
 
I assume the initial y-velocity is zero.

No other informtation other than what has been posted is given for the problem

Did I use the correct formula?
 
Can you let me know if my interpretation of the question is correct? If so, that would mean the bullet was shot out at some initial velocity, which it is true you don't know. However, because you know the acceleration acting on the bullet (9.8 m/s^2) at every point during its flight, and you know how long it stays in the air, those two things tell you exactly what vertical velocity the bullet needs to start off with.

If you can find that initial vertical velocity, and you know the total beginning velocity, which is 300 m/s, you can find the angle.
 
It's obvious that no one is able to solve this problem because they only ask questions (that lead nowhere) and all the information to the problem has been posted. Thank you all for your help anyway.
 

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