Finding Range of Object Shot at 40° Angle

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In summary, 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. The range of the object is 38.3 meters.
  • #1
Want to learn
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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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  • #2
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.
 
  • #3
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.
 
  • #4
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 :)
 
  • #5
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.
 

What is the formula for finding the range of an object shot at a 40° angle?

The formula for finding the range of an object shot at a 40° angle is R = (v^2 * sin(2θ)) / g, where R is the range, v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

What is the acceleration due to gravity used in the formula?

The acceleration due to gravity used in the formula is typically 9.8 m/s^2, which is the average value on Earth. However, this value may vary depending on the location and altitude.

Can the launch angle affect the range of the object?

Yes, the launch angle can significantly affect the range of the object. A launch angle of 40° is considered the optimal angle for maximum range, but any angle between 0° and 90° can be used.

What is the unit of measurement for the range?

The unit of measurement for the range will depend on the units used for velocity and acceleration. If the velocity is measured in meters per second (m/s) and the acceleration is measured in meters per second squared (m/s^2), then the range will be measured in meters (m).

Is air resistance considered in the calculation of range for objects shot at 40° angle?

No, the formula for finding the range of an object shot at a 40° angle does not take into account air resistance. This is because air resistance can vary depending on the shape, size, and material of the object, making it difficult to accurately calculate.

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