Projectile with fixed angle and tripled velocity

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SUMMARY

The discussion centers on a physics problem involving projectile motion, specifically a cannon with a fixed angle of projection that has an initial range of 1500 meters. When the initial speed of the cannonball is tripled, the correct calculation for the new range is derived from the physics of projectile motion, which states that range is proportional to the square of the initial velocity. Therefore, the new range is calculated as 1500 m multiplied by the square of 3, resulting in a range of 13500 m.

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enantiomer1
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Homework Statement


Hi, I'm stuck on this problem (it seems simple but I can't seem to get it down),
The question is, "A certain cannon with a fixed angle of projection has a range of 1500 m. What will be its range if you add more powder so that the initial speed of the cannonball is tripled?"


Homework Equations


d=vhori*t
d=vhori cos theta * t

The Attempt at a Solution


At first I simply saw this as a distance versus velocity and time problem (x=v0x*t so I simply tripled the distance getting 4500; not surprisingly, I was wrong. I'm sure that the equation relies on the 'fixed angle' part of the equation so I simply divided the equation by 21/2 (due to cos theta) getting 3192, but I'm still wrong, what variable am I forgetting?
 
Physics news on Phys.org
Welcome to PF.

If you triple the speed, what do you do to the horizontal velocity?

What does tripling the speed do to the time to max height and back down?

Distance is v * t so ...
 

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