Finding Range of Object Shot at 40° Angle

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Homework Help Overview

The problem involves determining the range of an object shot from the origin at a velocity of 50.0 m/s and an angle of 40.0 degrees above the horizontal, within the context of 2D motion equations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to calculate the flight time by first determining the time to reach the highest point and then the time to fall back to the ground. There is also mention of developing separate equations for the x and y components of motion and combining them to find the range.

Discussion Status

Some participants have provided guidance on how to approach the problem by focusing on the flight time and the relationship between the x and y components of motion. There is acknowledgment of missed steps in the original poster's reasoning, and a potential method has been suggested for combining equations.

Contextual Notes

The original poster expresses uncertainty about the sufficiency of the given information, particularly regarding the y-component of motion and the time variable.

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Homework Statement



An object is shot from the origin with a velocity of 50.0 m/s at an angle of 40.0 degrees above the horizontal. What is the range of the object?

Homework Equations


All the 2D Motion equations - too many to write all out. Ex:

[tex]V_x = V_0x + a_xt[/tex]


The Attempt at a Solution


I don't even think there is enough information given to solve this. I can only solve for
[tex]V_x[/tex] which is [tex]50 * cos 40 = 38.3[/tex] from there I don't know where to go. I try to figure out the y components of this motion, but we don't really know anything about y. We can't really find [tex]V_y[/tex] because to find that we need the equation [tex]V_y = v_0 sin \theta - gt[/tex] but we don't know the time.
 
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To work out the range we need to work out the flight time. So first find the time it takes the ball to reach it's highest point using the y-component of it's velocity. Then calculate how long it takes to hit the floor. The sum of these two times is the flight time.
 
Kalvarin said:
To work out the range we need to work out the flight time. So first find the time it takes the ball to reach it's highest point using the y-component of it's velocity. Then calculate how long it takes to hit the floor. The sum of these two times is the flight time.

ARRRGGGHHHHH - silly me. How could I miss that?

Thank you for pointing me in the right direction, much appreciated m8.
 
Want to learn said:
ARRRGGGHHHHH - silly me. How could I miss that?

Thank you for pointing me in the right direction, much appreciated m8.

Np :)
 
An simpler, more fruitful approach would be to develop two equations, and then turn them into one.

One would be [tex]x(t)[/tex], the other would be [tex]y(t)[/tex] and combining them will give you [tex]y(x)[/tex]

That function will describe the height of the object above the the origin as a function of its x-axis distance from the origin. Once you have that function, you can just set [tex]y=0[/tex] to find the x for which this holds true (The range.)

To get you started, I'll just rewrite your equation:

[tex]x(t)=v_0\cos{\theta}\cdot t[/tex]

[tex]t=\frac{x}{v_0\cos{\theta}}[/tex]

From here, you should be all set.
 

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