Time to Reach Maximum Height of Projectile Launched at 450 m/s

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To determine the time it takes for a bullet shot vertically at 450 m/s to reach maximum height, the formula v = u + at is used, with initial velocity (u) at 450 m/s and acceleration (a) as -10 m/s² or -9.81 m/s² depending on the calculation method. The first participant calculated 45 seconds using -10 m/s², while another participant arrived at 46 seconds using -9.81 m/s². Both calculations are valid, reflecting the slight difference in gravitational acceleration values. The discussion confirms the importance of understanding the impact of gravity on projectile motion. Overall, the time to reach maximum height is approximately 45 to 46 seconds, depending on the gravitational constant used.
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Hi...I done a question on velocity-time...I just wanted to amke sure if I am right...the question is: A soldier shoots a bullet vertically into the air at 450 m s^-1. How long does it take for bullet to reach maximum hight...air resistance can be ignored...so I used v=u+at, my final answer was 45 seconds..am I right...please reply anyone :rolleyes:
 
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Assuming you're using 10 m/s^2 for g, yes.
 
I second that. You're doing okay.
 
Thanks a lot.
 
dude i just did that same question for homework but i got 46s cos i used -9.81m/s2 for acceleration :D
 
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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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