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.

- idku

- katieb

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- FeeMeeGaming

AND?

- 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

http://i.imgur.com/I0prCLo.png
here is the IMAGE
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1) What is the initial speed of the ball?
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2) What is the initial angle θ of the ball with respect to the ground?
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3) What is the maximum height the ball goes above the ground?
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4) How far from where it was kicked will the ball land?
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5) What is the speed of the ball 2.5 seconds after it was kicked?
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6) How high above the ground is the ball 2.5 seconds after it is kicked?
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## More answers

- 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

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

- FeeMeeGaming

Sorry i dont
know.

- 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

`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

I am having trouble with part 5.

- 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

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

|dw:1441345370981:dw|

- idku

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

- idku

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

- idku

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

- anonymous

|dw:1441345702758:dw|

- idku

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

- idku

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

- anonymous

h is height from the ground=?

- idku

and what formula does this go by?

- idku

oh I think i got the time

- anonymous

h=ut+1/2gt^2

- anonymous

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

- 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

So at t=2.5 v=10.70m/s

- idku

12.70 ***

- idku

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

- idku

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

- idku

Yup got both time and max height

- idku

i mean not max height but height at t=2.5

- idku

Thank you very much!

- idku

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

- 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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