Projectile with fixed angle and tripled velocity

AI Thread Summary
The problem involves calculating the new range of a cannon when its initial speed is tripled while maintaining a fixed angle of projection. The original range is 1500 m, and the incorrect initial assumption was to simply triple this distance to 4500 m. The correct approach involves understanding that range is influenced by both the initial speed and the time of flight, which is affected by the angle of projection. Tripling the initial speed increases the horizontal velocity and affects the time to reach maximum height and return. The key takeaway is that the range increases by a factor of four when the speed is tripled, leading to a new range of 6000 m.
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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