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Projectile Kinematics

  • Thread starter Sylis
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  • #1
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Homework Statement


A rifle is aimed horizontally at a target 50m away. The bullet hits the target 2.5cm below the aim point.
A) Find the bullets flight time.
B)Find the bullets initial velocity upon leaving the barrel.

Homework Equations


Projectile kinematic equations/

The Attempt at a Solution



I've set it up as a triangle with the right angle on the right, the base of which is 50m and the height of which is 0.025m.

I understand how to find the hypotenuse but I'm not really sure what that tells me. Other than that I don't really feel like there's enough information to find out anything else. Thoughts?
 

Answers and Replies

  • #2
CAF123
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Begin by writing equations of motion for the horizontal and vertical directions.
 
  • #3
You have all the information you need. You're only given the distance in the x and y directions, the direction of the initial velocity (horizontal in positive x direction). Everything else is known, like the downward acceleration due to gravity which is [itex]9.8m/s^{2}[/itex].

It sounds like you've set up one large triangle to span the entire path of the projectile, and it also sounds like the hypotenuse is what you think is the path of the projectile. You're on the right track with using the triangles, although you usually use a triangle for each moment being measured. Such as the initial velocity/direction, and the final velocity/direction. Also, projectile motion does not follow a straight line, it's curved like an upside down parabola - not the hypotenuse of a triangle.

The first thing to do in these problems is lay out all the Kinematic equations and see what you known and don't have yet. (hint: a lot of terms are zero'ed out)

Motion in x direction:
[itex]x_{1}=x_{0}+ v_{x0}t+\frac{1}{2}a_{x}t^2[/itex]

Motion in y direction:
[itex]y_{1}=y_{0}+v_{y0}t+\frac{1}{2}a_{y}t^2[/itex]

[itex]v_{y1}=v_{y0}+a_{y}t[/itex]

[itex]v_{y1}^2=v_{y0}^2+2a_{y}(y_{1}-y_{0})[/itex]

If you write these four equations on your sheet for every projectile motion problem, it will become easy work.

The first thing I would do is the last equation, because you have everything except one unknown [itex]v_{y1}[/itex] (which is the final velocity in the negative y direction).

[itex]v_{y1}=\sqrt{2a_{y}(y_{1}-y_{0})}[/itex]

[itex]y_{0}[/itex] is zero, and [itex]y_{1}[/itex] is 0.025cm (and I believe we're looking for the ABS of this since it's in a sqrt, someone correct me pleaseeee) if you think only about the y direction of this problem, it's just like dropping a ball (bullet) from the height, and seeing how long it takes to drop 2.5cm. The x direction part is seprate. If you find how fast the bullet drops 2.5cm, that can be used to find the airtime, which can also tell you the initial velocity in the x direction.

Don't forget to convert 2.5cm to 0.025m!

[itex]v_{y1}=\sqrt{2(9.8)(0.025)} = 0.7m/s[/itex]

So there's one unknown, and three other equations to work with.

disclaimer: i'm a student on the same level, posting here to better refine my understanding of the topic, if any more experienced members decide this is all totally wrong information, please let me know!
 
  • #4
CAF123
Gold Member
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88
Motion in y direction:
[itex]y_{1}=y_{0}+v_{y0}t+\frac{1}{2}a_{y}t^2[/itex]

[itex]v_{y1}=v_{y0}+a_{y}t[/itex]

[itex]v_{y1}^2=v_{y0}^2+2a_{y}(y_{1}-y_{0})[/itex]

The first thing I would do is the last equation, because you have everything except one unknown [itex]v_{y1}[/itex] (which is the final velocity in the negative y direction).
If you want the bullets flight time, then the first kinematic equation in your list gives you that right away.
[itex]y_{0}[/itex] is zero, and [itex]y_{1}[/itex] is 0.025cm (and I believe we're looking for the ABS of this since it's in a sqrt, someone correct me pleaseeee)
Displacement and acceleration are vector quantities. You defined downwards as positive. You could have equally as well defined it to be negative, in which case ##a_y = -9.8## and ##y_1 - y_0 = -0.025##. So their product is still positive under sqrt. But then of course, you have to interpret the sign of ##v_{y_1}## correctly.
 

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