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idku
 one year ago
Math  Based  Physics .... QUESTIONS.
(I am working the parts 16 on my own, if you want, please come and view and correct if anything)
A soccer ball is kicked with an initial horizontal velocity of 11 m/s and an initial vertical velocity of 17 m/s.
idku
 one year ago
Math  Based  Physics .... QUESTIONS. (I am working the parts 16 on my own, if you want, please come and view and correct if anything) A soccer ball is kicked with an initial horizontal velocity of 11 m/s and an initial vertical velocity of 17 m/s.

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idku
 one year ago
Best ResponseYou've already chosen the best response.1I got disconnected I apologize. I will post parts of the problem will do them here on my own and if I need help i will look for it.

idku
 one year ago
Best ResponseYou've already chosen the best response.1http://i.imgur.com/I0prCLo.png here is the IMAGE  1) What is the initial speed of the ball?  2) What is the initial angle θ of the ball with respect to the ground?  3) What is the maximum height the ball goes above the ground?  4) How far from where it was kicked will the ball land?  5) What is the speed of the ball 2.5 seconds after it was kicked?  6) How high above the ground is the ball 2.5 seconds after it is kicked? 

idku
 one year ago
Best ResponseYou've already chosen the best response.1` PART 1: ` The initial speed of the ball (lets denote it with u) is: \(\large u=\sqrt{11^2+17^2}=20.24\color{blue}{\left({\rm m/s}\right)}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.1` PART 2:` \(\large \theta={\rm Tan}^{1}\left({\rm 17/11}\right)\approx 57.09\)

FeeMeeGaming
 one year ago
Best ResponseYou've already chosen the best response.0Sorry i dont know.

idku
 one year ago
Best ResponseYou've already chosen the best response.1Now for part 3, we know that maximum height is reached when v=0. Also, I am given: a = 9.81m/s\(^2\) u = 20.24m/s So will I use? \(v^2=u^2+2ax\) NOTE: "u" denotes initial speed. "x" denotes displacement/distance (in this case same thing) \(v^2=u^2+2ax\) \(0^2=17^2+2(9.81)x\) \(17^2=2(9.81)x\) \(17^2=2(9.81)x\) \(x=17^2/[2(9.81)]\approx 14.73\)

idku
 one year ago
Best ResponseYou've already chosen the best response.1`PART 4` How far from where it was kicked will the ball land? `(Horizontal Displacement) = (Time) × (Horizontal Velocity)` I know the horizontal velocity, but the time is unknown. We can use the vertical components to find the time!! Since the ball travels in a parabolic motion it is just twice the time from v=17 (initial horiz velocity) to v=0 (the maximum height point, or the vertex) so we went from 17 m/s to 0 m/s how many second would that have took if every second we lose 9.81m/s? 17(m/s) / 9.81(m/s^2) ≈ 1.732 seconds Now remember we go twice from v=17m/s to v=0m/s So we really have 3.464 seconds of FULL motion time when ball is in air. units work out too! Now we can find the time. `(Horizontal Displacement) = (Time) × (Horizontal Velocity)` `(Horizontal Displacement) = (3.464) × (11)` `(Horizontal Displacement) = 38.104`

idku
 one year ago
Best ResponseYou've already chosen the best response.1I am having trouble with part 5.

idku
 one year ago
Best ResponseYou've already chosen the best response.1A soccer ball is kicked with an initial horizontal velocity of 11 m/s and an initial vertical velocity of 17 m/s. 5) What is the speed of the ball 2.5 seconds after it was kicked?

idku
 one year ago
Best ResponseYou've already chosen the best response.1v=u+at u  initial speed t  time a  accel due to gravity = 9.81m/s^2 v=20.24+(9.81)2.5=4.285 speed is always positive, so 4.285 because we want to know what happens with speed at t=2.5 No, that doesn;t work for some reason.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441345370981:dw

idku
 one year ago
Best ResponseYou've already chosen the best response.111+(9.81)2.5=35.525 and that is incorrect either (I put it in my practice thingy and it denied)

idku
 one year ago
Best ResponseYou've already chosen the best response.1oh, 20.24 not 11, I suppose. initial velocity V< not speed of horiz.

idku
 one year ago
Best ResponseYou've already chosen the best response.120.24+(9.81)2.5=44.675 but that is too large. and I check it is not right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1441345702758:dw

idku
 one year ago
Best ResponseYou've already chosen the best response.1I am sorry but I don't really get that picture.

idku
 one year ago
Best ResponseYou've already chosen the best response.10.768 second from 2.5 to the vertx point. g=9.81 and what is the h?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0h is height from the ground=?

idku
 one year ago
Best ResponseYou've already chosen the best response.1and what formula does this go by?

idku
 one year ago
Best ResponseYou've already chosen the best response.1oh I think i got the time

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0t=1.7320.768=? dw:1441346465866:dw

idku
 one year ago
Best ResponseYou've already chosen the best response.1v=u+at but we are starting from v=0 using t=0.768 because there is 0.768 seconds till the time point of 2.5s that we want. (This is making sense a lot, if we just flip the pic from left to right) So v=20.24+(9.81m/s^2)(0.768s)=12.70m/s

idku
 one year ago
Best ResponseYou've already chosen the best response.1h=ut+1/2gt^2 are we using the vertical velocity component 17m/s for u?

idku
 one year ago
Best ResponseYou've already chosen the best response.1Vertical Displacement = (17m/s)•(2.5s) + (1/2)•(9.81m/s\(^2\))•(2.5s)\(^2\)=11.84m

idku
 one year ago
Best ResponseYou've already chosen the best response.1Yup got both time and max height

idku
 one year ago
Best ResponseYou've already chosen the best response.1i mean not max height but height at t=2.5

idku
 one year ago
Best ResponseYou've already chosen the best response.1\(x=ut+\dfrac{1}{2}(9.81)t^2\) x  vert. displacement u  initial velocity

idku
 one year ago
Best ResponseYou've already chosen the best response.1that is a nice formula I am taking a note of. Thanks once again. (((I will do a similar practice once more.)))
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