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idku
 one year ago
Please help me with 2 brief questions.
idku
 one year ago
Please help me with 2 brief questions.

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idku
 one year ago
Best ResponseYou've already chosen the best response.1A cannonball is shot (from ground level) with an initial horizontal velocity of 33 m/s and an initial vertical velocity of 25 m/s.  \(\rm \large I~found~so~far:\) [1] initial speed, \(\rm u=41.4(m/s)\) [2] angle of cannonball with respect to ground, \(\rm \theta=37.15º\) [3] maximum height the cannonball goes above the ground \(\rm x=31.8(m)\) [4] horizontal displacement is \(\rm 167.6(m)\) [5] the entire time of the ball motion \(\rm 5.1(s)\)  \(\rm \large I~need:\) [6] speed of the cannonball at, \(\rm t=3.7(s)\) [7] height (Vertical Displacement) at, \(\rm t=3.7(s)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.06.Vi = √[Vx²+Vy²] = 44.2 m/s 7.3) Hmax = Vi²sin²Θ/(2g) = 62.49 m

idku
 one year ago
Best ResponseYou've already chosen the best response.1Are you sure? Becsause that doesn't seem to be right.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Don't know,I never liked physics. Maybe he can help @Hero @dan815 @iambatman

idku
 one year ago
Best ResponseYou've already chosen the best response.1I found the vertical displacement x=25.35m applying: \(x = ut + \dfrac{1}{2}at^2\) I have put: t=3.7s a=9.81m/s\(^2\) u=25m/s

idku
 one year ago
Best ResponseYou've already chosen the best response.1I need velocity at t=3.7 the rest i found. tnx for coming btw

dan815
 one year ago
Best ResponseYou've already chosen the best response.1A cannonball is shot (from ground level) with an initial horizontal velocity of 33 m/s and an initial vertical velocity of 25 m/s. [6] speed of the cannonball at, t=3.7(s) [7] height (Vertical Displacement) at, t=3.7(s)

dan815
 one year ago
Best ResponseYou've already chosen the best response.1speed is the magnitude of velocity

dan815
 one year ago
Best ResponseYou've already chosen the best response.1velocity tells you how speed is broken into components of direction

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(\large V_y=259.81(3.7)=11.297\) \(\large V_x=33\) (yes, it doesn't change b/c acceleration is 0) \(\large V_t=\sqrt{33^2+11.297^2}=34.88\)

idku
 one year ago
Best ResponseYou've already chosen the best response.1oh, that is what I wanted to do. to use pythagorean theoream after fiding vert and horiz velocities (well horiz veloc I knew...:D) I was getting NEGATIVE 11.297 so I was thinking it is wrong.

idku
 one year ago
Best ResponseYou've already chosen the best response.1I mean those veloc at t=3.7

dan815
 one year ago
Best ResponseYou've already chosen the best response.1negative velocity just means down, it an indication of direction thats all

idku
 one year ago
Best ResponseYou've already chosen the best response.1I will take a note of that in my brain. Ok, so negative velocity is also an acceptable result (unlike displacement)

dan815
 one year ago
Best ResponseYou've already chosen the best response.1to someone who flipped the page, he will think the Veocity is positive it doesnt matter to him its the way u chose your coordinates

idku
 one year ago
Best ResponseYou've already chosen the best response.1I want to flip the page, but another page. The page of the physics intersecting w/ my life.... I don';t even know what I would do without you

dan815
 one year ago
Best ResponseYou've already chosen the best response.1get used to it, uve intersected with physics from the day u were born

idku
 one year ago
Best ResponseYou've already chosen the best response.1yes, but not with due dates

idku
 one year ago
Best ResponseYou've already chosen the best response.1and not with rediculous teachers such as my prof at college

idku
 one year ago
Best ResponseYou've already chosen the best response.1In any case, I guess  Alright. I have a report on a lab. That I should be able to do. (Got 6 hours to work on it before my next classes)

idku
 one year ago
Best ResponseYou've already chosen the best response.1Have a good night or where ever yo are. Niagara falls:D
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