A soccerball was kicked at an angle of 28 deg to the ground and it landed 39.8m away.
a) what was the velocity of the ball when kicked?

- YumYum247

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- katieb

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- YumYum247

@Astrophysics just give me a quick hint and leave the rest to me.....

- Astrophysics

VECTORS PROJECTILE MOTION YAYAYAYA DIAGRAMS

- YumYum247

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## More answers

- Astrophysics

|dw:1439959033896:dw| I wanted to make an arc

- YumYum247

just tell me what's the first thing i need to look for....that's all

- Astrophysics

|dw:1439959053044:dw| now lets make it nice and clean so we know exactly what's going on, we need to find v0 here which is velocity of the ball when it was kicked.

- Astrophysics

You need to find your velocities at x and y direction, play around with it and apply kinematic formulas

- YumYum247

yes but for that i need to find the final velocity :(

- YumYum247

|dw:1439959268885:dw|

- Astrophysics

We don't actually need it, I have an idea!!!!

- Astrophysics

Want to hear it

- YumYum247

yuss!!!!

- YumYum247

Vertical: Horizontal:
a = -9.8m/sec2 dh = 39.8m
Dh = 0m/s Vh = ?
Vi = ? t = ?
Vf = ?
t = ?

- Astrophysics

Ok I was actually thinking it gave us the maximum height

- YumYum247

nope that's all we got???

- Astrophysics

Good lets put everything in x and y direction...ok lets see and thing and have fun!

- YumYum247

sorry i forgot to add the theta in there...
Vertical: Horizontal:
a = -9.8m/sec2 dh = 39.8m
Dh = 0m/s Vh = ?
Vi = ? t = ?
Vf = ?
t = ?
Angle = 28deg

- Astrophysics

x - direction
x = 39.8 m
\[\large v_{x_0} = v_0 \cos (28)\]
y-direction
a = -g
\[\large v_{y_0} = v_0 \sin(28)\]
so this is what we're given hahahaaa

- YumYum247

yes....

- Astrophysics

Ok ok we can actually use the range formula

- Astrophysics

and find initial velocity instantly XD

- Astrophysics

\[R = \frac{ v_0^2\sin(2 \theta) }{ g }\]

- YumYum247

hey about that..... what does it mean by Vi here, is it the horizontal velo or the vertical???

- Astrophysics

Oh no that's the resultant velocity |dw:1439960011427:dw| what we're looking for

- Astrophysics

That can be split into components as shown to figure it out

- YumYum247

yah cuz that formula says dh = Vi^2 X Sin2theta/a
so that's why i was wondering what does that initial velocity means here....

- YumYum247

horizontal or vertical...cuz if it's the resultant then shouldn't it be Vf^2????

- Astrophysics

I don't follow, that's the initial velocity, we are looking for the velocity that it was kicked at not at the velocity it landed at

- YumYum247

|dw:1439960370291:dw|

- Astrophysics

Oh I see what you're trying to do, no that's not right, because we're splitting the initial velocity into components

- Astrophysics

|dw:1439960455619:dw|

- Astrophysics

Those are the components of the initial velocity, does that make sense?

- Astrophysics

you can label it as i for initial it's the same thing :P

- YumYum247

those nots confuse me ......i only understand Vf, Vi and Vh

- Astrophysics

Ok sorry ill fix it

- YumYum247

Thanks :")

- Astrophysics

|dw:1439960585668:dw| same thing

- YumYum247

oh now i see...... :D

- Astrophysics

Yeah they are just components of the initial velocity :-)

- Astrophysics

\[R = \frac{ v_i^2\sin(2 \theta) }{ g }\]

- YumYum247

ok so where are we going with that formula, are we solving for the Vi there???

- Astrophysics

Yup!

- YumYum247

ok so it'll be like this....

- YumYum247

|dw:1439960730234:dw|

- YumYum247

do we take the square root??? O_o

- Astrophysics

Mhm hold on a sec, lets just do the algebra first to avoid confusion the range here is our x or distance which is 39.8 m.
It's always best to do algebra first then plug things in especially when you get into deeper physics, it will benefit you.
\[x= \frac{ v_i^2 \sin(2 \theta) }{ g }\] now solve for the initial velocity

- Astrophysics

Don't plug anything in just solve for initial velocity

- Astrophysics

variable

- YumYum247

ok so rearrange the variables first....

- Astrophysics

Yup

- YumYum247

and then plug i nthe numbers....???

- Astrophysics

Yeah

- YumYum247

aight thanks man, i really appreciate your help..... Stay cute and blessed!!!! :")

- Astrophysics

Np, lets just finish it off :P, so what do you get when you solve for initial velocity when you rearrange don't worry about making mistakes, that's how we learn!

- YumYum247

hold up.....

- Astrophysics

Take your time

- YumYum247

x = Vi^2 X Sin2theta/a
x.a = Vi^2 X Sin2theta
x.2/sin2Theta = Vi^2

- Astrophysics

Perfect now just square root it and we're good!

- YumYum247

|dw:1439961336297:dw|

- Astrophysics

I'm assuming that 2 in the numerator was meant to be an a haha

- YumYum247

ok man thanks alot i really appreciate your effort.... stay blessed!! :)

- Astrophysics

|dw:1439961409567:dw| whats that

- YumYum247

x

- YumYum247

or R

- YumYum247

39.8m

- Astrophysics

Hmm...our equation was \[x= \frac{ v_i^2 \sin(2 \theta) }{ g } \implies v_i = \sqrt{\frac{ gx }{ \sin(2 \theta) }}\]

- Astrophysics

I think you may have added an extra x by mistake

- YumYum247

yah!!! thats it....!!!

- YumYum247

no that's just my hand writing :"D

- YumYum247

ok thanks and good night, sleep tight and don't let the bugs bite :"D

- Astrophysics

Take care :)

- YumYum247

too 2 :)

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