Projectile with fixed angle and tripled velocity

[enantiomer1]

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=v_hori*t
d=v_hori cos theta * t

The Attempt at a Solution

At first I simply saw this as a distance versus velocity and time problem (x=v_0x*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 2^1/2 (due to cos theta) getting 3192, but I'm still wrong, what variable am I forgetting?

 

[LowlyPion]

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 ...

 

[thomate1]

I would first make sure that all the units are consistent and in the correct form. In this case, the initial velocity of the cannonball should be in meters per second (m/s) and the time should be in seconds (s).

Then, I would use the given formula d=vhori*t to calculate the initial velocity (vhori) of the cannonball. Since the range is given as 1500 m, we can rearrange the formula to solve for vhori:

vhori = d/t = 1500 m/ t

Now, we know that the initial velocity is directly proportional to the range, so if we triple the initial velocity, the range should also triple. Therefore, the new range would be 3 times the original range, which is 3*1500 = 4500 m.

In summary, the range of the cannonball would be 4500 m if the initial velocity is tripled. It is important to note that this calculation assumes that the angle of projection remains fixed and the air resistance is negligible.