Solve Projectile Motion Problems with Tripled Speed | Physics Help"

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A projectile launched at an angle lands on the same horizontal plane with a range of 23m. When launched again at triple the speed, the new range needs to be calculated. The user is unsure about using the equation x = v0t to find time and how to incorporate acceleration into the range calculation. They plan to use the equation x = v0t + 1/2at^2 but seek confirmation on their approach. Forum members emphasize the importance of following the homework template and providing relevant equations for effective assistance.
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Physics problem helpp!

1. A projectile is launched at a angle and it lands on the same horizontal plane. The range of this launch is 23m. The projectile is launched again at the same angle with triple the speed. What is the new range?

2. x=v0t <---- I know that finding the time would be useful, but i am not sure if this is the right equation to solve for time.

3. I know that x= 23m but i have no idea how to find the new range. After I find out what the time is I would triple it and plug it into the equation x= v0t+1/2at^2. I'm not sure if this is right!
Thank You if you can help me.
 
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Hi tomhall, welcome to PF. We can help you, but you need to follow the rules of the forum. You need to use the homework template, show the relevant equations and tell us what you think you should do with them.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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