What is the horizontal range of a projectile fired into the air?

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SUMMARY

The horizontal range of a projectile fired into the air can be calculated using the initial velocity components derived from the velocity vector v = (8.6i + 5.0j) m/s observed at 3.3 seconds. The horizontal component of the velocity, Vx, is 8.6 m/s, and the time of flight is 3.3 seconds. The horizontal range R is determined using the formula R = Vx * t, resulting in a calculated range of 28.38 meters. The initial misunderstanding regarding the relationship between R and the variables has been clarified, emphasizing the importance of using the correct components of the velocity vector.

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Homework Statement


Exactly 3.3s after a projectile is fired into the air from the ground, it is observed to have a velocity v = (8.6i + 5.0j ), where the x-axis is horizontal and the y-axis is positive upward.

Determine the horizontal range of the projectile

Homework Equations


x=x0+vy0t+1/2at2
v=[tex]\sqrt{}V<sub>x</sub>[/tex]2+Vy2

The Attempt at a Solution


v= 9.95 by finding the resultant of the velocity vector
x 0 = 0
ax= 0
x= Vx(t)?
is that correct?
if so then,
x= 9.95(3.3)= 32.8?
but this is wrong and i don't know how to get it
wait does R= x? or y?
 
Last edited:
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Use the information that v = (8.6i+5.0j)m/s in 3.3 s to find the components of the initial velocity vector. Then use the equation that relates the range R to these initial components to answer the question.
 

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