idku
  • idku
Math - Based - Physics .... QUESTIONS. (I am working the parts 1-6 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.
Mathematics
katieb
  • katieb
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FeeMeeGaming
  • FeeMeeGaming
AND?
idku
  • idku
I 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
  • idku
http://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? ------------------------------------------------------------

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idku
  • idku
` 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
  • idku
` PART 2:` \(\large \theta={\rm Tan}^{-1}\left({\rm 17/11}\right)\approx 57.09\)
FeeMeeGaming
  • FeeMeeGaming
Sorry i dont know.
idku
  • idku
Now 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
  • idku
`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
  • idku
I am having trouble with part 5.
idku
  • idku
A 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
  • idku
v=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
  • anonymous
|dw:1441345370981:dw|
idku
  • idku
11+(9.81)2.5=35.525 and that is incorrect either (I put it in my practice thingy and it denied)
idku
  • idku
oh, 20.24 not 11, I suppose. initial velocity V< not speed of horiz.
idku
  • idku
20.24+(9.81)2.5=44.675 but that is too large. and I check it is not right
anonymous
  • anonymous
|dw:1441345702758:dw|
idku
  • idku
I am sorry but I don't really get that picture.
idku
  • idku
0.768 second from 2.5 to the vertx point. g=-9.81 and what is the h?
anonymous
  • anonymous
h is height from the ground=?
idku
  • idku
and what formula does this go by?
idku
  • idku
oh I think i got the time
anonymous
  • anonymous
h=ut+1/2gt^2
anonymous
  • anonymous
t=1.732-0.768=? |dw:1441346465866:dw|
idku
  • idku
v=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
  • idku
So at t=2.5 v=10.70m/s
idku
  • idku
12.70 ***
idku
  • idku
h=ut+1/2gt^2 are we using the vertical velocity component 17m/s for u?
idku
  • idku
Vertical Displacement = (17m/s)•(2.5s) + (1/2)•(-9.81m/s\(^2\))•(2.5s)\(^2\)=11.84m
idku
  • idku
Yup got both time and max height
idku
  • idku
i mean not max height but height at t=2.5
idku
  • idku
Thank you very much!
idku
  • idku
\(x=ut+\dfrac{1}{2}(-9.81)t^2\) x - vert. displacement u - initial velocity
idku
  • idku
that 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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