## idku one year ago 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.

1. FeeMeeGaming

AND?

2. 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.

3. 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? ------------------------------------------------------------

4. idku

 PART 1:  The initial speed of the ball (lets denote it with u) is: $$\large u=\sqrt{11^2+17^2}=20.24\color{blue}{\left({\rm m/s}\right)}$$

5. idku

 PART 2: $$\large \theta={\rm Tan}^{-1}\left({\rm 17/11}\right)\approx 57.09$$

6. FeeMeeGaming

Sorry i dont know.

7. 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) $$v^2=u^2+2ax$$ $$0^2=17^2+2(-9.81)x$$ $$-17^2=2(-9.81)x$$ $$17^2=2(9.81)x$$ $$x=17^2/[2(9.81)]\approx 14.73$$

8. 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

9. idku

I am having trouble with part 5.

10. 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?

11. 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.

12. anonymous

|dw:1441345370981:dw|

13. idku

11+(9.81)2.5=35.525 and that is incorrect either (I put it in my practice thingy and it denied)

14. idku

oh, 20.24 not 11, I suppose. initial velocity V< not speed of horiz.

15. idku

20.24+(9.81)2.5=44.675 but that is too large. and I check it is not right

16. anonymous

|dw:1441345702758:dw|

17. idku

I am sorry but I don't really get that picture.

18. idku

0.768 second from 2.5 to the vertx point. g=-9.81 and what is the h?

19. anonymous

h is height from the ground=?

20. idku

and what formula does this go by?

21. idku

oh I think i got the time

22. anonymous

h=ut+1/2gt^2

23. anonymous

t=1.732-0.768=? |dw:1441346465866:dw|

24. 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

25. idku

So at t=2.5 v=10.70m/s

26. idku

12.70 ***

27. idku

h=ut+1/2gt^2 are we using the vertical velocity component 17m/s for u?

28. idku

Vertical Displacement = (17m/s)•(2.5s) + (1/2)•(-9.81m/s$$^2$$)•(2.5s)$$^2$$=11.84m

29. idku

Yup got both time and max height

30. idku

i mean not max height but height at t=2.5

31. idku

Thank you very much!

32. idku

$$x=ut+\dfrac{1}{2}(-9.81)t^2$$ x - vert. displacement u - initial velocity

33. idku

that is a nice formula I am taking a note of. Thanks once again. (((I will do a similar practice once more.)))