How can I find the horizontal distance without knowing the initial velocity?

[kevinlikesphysics]

how do i solve for this i only know how to do it with a degree and an Vi

the question is

a football is kicked at 45 degrees and travels 120 m before hitting the ground

and i need

Vi
time in air
and max height

how do i get vcos45 without a velocity?

thanks

 

[Fermat]

If you did know Vi and an angle, could you work out how far it traveled ?
How would you do that ?

 

[kevinlikesphysics]

i don't have Vi


edit: could you work it out for me like with just equations

 

[Tide]

i don't have Vi
edit: could you work it out for me like with just equations

I think the rules here are that you cannot get direct help on a homework problem until you have at least shown some of your own work.

 

[Fermat]

i don't have Vi
edit: could you work it out for me like with just equations

You have to find Vi, yes,
But, if if you did have a value for Vi, and also had an angle, could you then work out the distance travelled.
Just use symbols and show us some working.

 

[Beam me down]

how do i get vcos45 without a velocity?
thanks

This should help you along a bit:


V(y)=vsin45
V(x)=vcos45

So t=(2vsin45)/g

d(x)=120

d(x)=v(x)t
120=(vcos45)((2vsin45)/g)
120g=2vcos45vsin45
v^2=(120g)/(2cos45sin45)
v=sqrt{(120g)/(2cos45sin45)}

 

[lawtonfogle]

This should help you along a bit:
V(y)=vsin45
V(x)=vcos45
So t=(2vsin45)/g
d(x)=120
d(x)=v(x)t
120=(vcos45)((2vsin45)/g)
120g=2vcos45vsin45
v^2=(120g)/(2cos45sin45)
v=sqrt{(120g)/(2cos45sin45)}

I think that constitutes giving him to much help. I think there is a equation that makes it simpilar, I'm looking now.
Ok, found it. Now, let's hope I remember how to layTex.
$$R = (V^2_0 \cdot 2 \cdot sin \theta) / g$$
this is the formula you use to find $$V_0$$ or as some say, $$V_i$$

 

[lawtonfogle]

I kinda remember. I want it to put the multipling dots, but instead it puts dots on top. Hmmm... Guess I will go find out by looking at other tex.

 

[NoMeGusta]

I think that constitutes giving him to much help. I think there is a equation that makes it simpilar, I'm looking now.
Ok, found it. Now, let's hope I remember how to layTex.
$$R = (V^2_0 \cdot 2 \cdot sin \theta) / g$$
this is the formula you use to find $$V_0$$ or as some say, $$V_i$$

is R the distance ?

 

[lawtonfogle]

Yes, it is the distance the object has traveled.

 

[Tide]

It is the horizontal distance the object has traveled. :)