Very vague question about projectile launch angle

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SUMMARY

The discussion centers on determining the launch angle of a projectile given that its launch speed is 5.2 times its speed at maximum height. Key insights include the understanding that at maximum height, the vertical component of velocity is zero, leaving only the horizontal component. The relationship between the launch speed and the horizontal velocity can be established using trigonometric principles. This provides a clear pathway to calculate the launch angle using the known speed ratio.

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The launching speed of a certain projectile is 5.2 times the speed it has at maximum height. What is the elevation angle at launching?

Could someone just give me a starting point or a hint in the right direction?

I know that the horizontal velocity doesn't decrease, and the vertical velocity is 0 at the maximum height. I can't put it together though; it seems like there isn't enough information given. How can you possibly tell by how much of a factor the speed is being affected by the vertical velocity, when you don't know the horizontal velocity?
 
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Maybe you can assume the magnitude [tex]V[/tex] of the lunching speed and find the relation between it and the horizontal velocity.Hint: use trigonometry
 
Last edited:
At the maximum height the y component of the velocity is zero, therefore the speed at maximum height is only the x-component of the motion. The x-component of the velocity stays the same throughout the motion.
 

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