# Non-Horizontally launched Projectile

physics(L)10

## Homework Statement

A ball is thrown from the ground (height=0) with an initial speed Vo at an angle of 45 degrees

a)when does ball start falling to earth
b)what is balls max height
c)How long does it take to hit the ground again?
d)where does the ball hit the ground?

## Homework Equations

x=Xo + Uxt + (at^2)/2
z=Zo + Uxt + (at^2)/2

r=(0,0,0)
u=Vo(cos45,0,sin45)
a=(0,0,-g)

## The Attempt at a Solution

x= Ut
z=cos45t + (sin45t^2)/2

Now I'm stuck lol

Homework Helper
Well when the ball starts falling back down, what should the vertical velocity be?

physics(L)10
Yeah, I know that, velocity=0 right? but which equation do I make =0? what do I do after this?

Homework Helper
Yeah, I know that, velocity=0 right? but which equation do I make =0? what do I do after this?

Well you know that v2=v02+2as, v=v0+at as well as the equation you put. So which one will give you time when v=0?

physics(L)10
We don't have a value for Vo though. That's what's confusing me.

Homework Helper
In that case, I suspect you should find your answers in term of v0

physics(L)10
Wtf, how would you do that?

Homework Helper
Wtf, how would you do that?

You have vx=v0cos45 and vy=v0sin45. So when considering horizontal motion, you can write

x=vxt as x=(v0cos45)t

Like that.

physics(L)10
So the only way to do it is to have an unknown variable still there?

Homework Helper
Yes, you were not given enough information to find v0

physics(L)10
So 0= t because anything divided by 0 is 0, correct?

Homework Helper
So 0= t because anything divided by 0 is 0, correct?

Where did you get that from?

You have three motion equations

s=ut+1/2at2
v2=u2+2as
v=u+at

For the first, part, which one will give you time when the final vertical velocity is zero?

Maybe_Memorie
So 0= t because anything divided by 0 is 0, correct?

Not quite.
It's undefined. Plug a number divided by 0 into your calculatior. It'll say Math Error or something like that..