Recent content by Milkster18
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Projectile Motion (Initial Velocity)
Ah W1 + W2 / 2 20 + 40 / 2 = 30 rpm So now i have average how does this make things easier? Im from the UK by the way so i think some units / symbols are different to US if your from there. Thanks for the help again. :)- Milkster18
- Post #33
- Forum: Introductory Physics Homework Help
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Projectile Motion (Initial Velocity)
one revolution = 2*pi radians 31.5 / 2*pi = just over 5 revolutions :) but i had to find 2*pi first then use that answer to divided into 31.5 rads.. otherwise the answer will be completely different for some reason. As for average angular speed: 20 / 2 = 10 rpm?- Milkster18
- Post #31
- Forum: Introductory Physics Homework Help
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Projectile Motion (Initial Velocity)
31.5 radians.. my mistake- Milkster18
- Post #28
- Forum: Introductory Physics Homework Help
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Projectile Motion (Initial Velocity)
So use equation Pheta = W1 + 0.5at^2 W1 = 2.1rad/s W2 = 4.19 rad/s Pheta = 2.1(10) + 0.5(0.21)(10)^2 Pheta = 21 + 10.5 Pheta = 30.5 radians? How do i use this to find revolutions if this is correct?- Milkster18
- Post #27
- Forum: Introductory Physics Homework Help
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Projectile Motion (Initial Velocity)
Ok looking good. Square root 8.86^2 + 22.22^2 =23.9 m/s Ui = 23.9 m/s .. looking good its correct :) Any reason to find the angle? I am guessing to help find other components? If anything ever comes to using quadratic equations is there any other method that is easier such as the average...- Milkster18
- Post #25
- Forum: Introductory Physics Homework Help
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Projectile Motion (Initial Velocity)
Ok, sorry i didn't know about distance = avg speed x time formula. Wasn't told by my lecturer about that. yes distance / speed sounds better than speed / distance lol. so now i find Ux given by time. Sx = Ux t Ux = Sx / t = Ux = 20 / 0.9 = Ux as 22.22 m/s so if I am finding average of Ux...- Milkster18
- Post #23
- Forum: Introductory Physics Homework Help
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Projectile Motion (Initial Velocity)
Ok so in order to find t i use: Ur saying: Distance = avg speed x time So time = distance / avg speed? Got 2 distances sy = 4, sx = 20m Not sure wat to do here.. Im guessing 4 / 4.43 = 0.9s Or 4.43/4 = 1.11s- Milkster18
- Post #21
- Forum: Introductory Physics Homework Help
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Projectile Motion (Initial Velocity)
Yes i realize using quadratics is a way to find time, i didn't know there was an easier way to solve for t thou. as for getting the average vertical initial velocity .. i don't know what this answer goes into which equation in order to solve t. Sorry i haven't solved for initial velocity...- Milkster18
- Post #19
- Forum: Introductory Physics Homework Help
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Projectile Motion (Initial Velocity)
Question: Why does the AVERAGE Initial vertical velocity have to be used instead of just the initial vertical velocity so i can understand why?- Milkster18
- Post #16
- Forum: Introductory Physics Homework Help
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Projectile Motion (Initial Velocity)
Sy = (4.43)t - (4.91)t^2 4 = (4.43)t - (4.91) 4 = -0.48? 4 / 0.48 = square root 8.33 t = 2.887 s ?- Milkster18
- Post #15
- Forum: Introductory Physics Homework Help
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Projectile Motion (Initial Velocity)
with the analogy: Say you made 0$ to 100$ over 7 days .. average income per day is 100 / 7 = 14.3$ per day- Milkster18
- Post #14
- Forum: Introductory Physics Homework Help
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Projectile Motion (Initial Velocity)
u+v / 2 = the average? So v = 0 8.86 / 2 = 4.43 m/s- Milkster18
- Post #12
- Forum: Introductory Physics Homework Help
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Projectile Motion (Initial Velocity)
sice it highest point is Sy = 4m is it 8.86 / 4 = 2.215 m/s average vertical speed? if its not can you provide further help? thanks- Milkster18
- Post #10
- Forum: Introductory Physics Homework Help
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Projectile Motion (Initial Velocity)
speed is calculating in m/s.. sorry I am confused- Milkster18
- Post #8
- Forum: Introductory Physics Homework Help
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Projectile Motion (Initial Velocity)
Average vertical speed.. Uy / g 8.86 / 9.81 = 0.9 s? (Sorry I am getting tired as its nearly 2am)- Milkster18
- Post #7
- Forum: Introductory Physics Homework Help