Find the equation for horizontal motion

[clintau]

a ball is thrown with v(initial)=5i+19.6j

find the equation of the horizontal motion

Homework Equations

the problems asks for ymax and tymax and part b asks for the equations of the horizontal and vertical motions.

Lost on the equations

 

[Kurdt]

What have you tried so far?

 

[shamrock5585]

your velocity is a vector... break it into components and you will see that your horizontal vector is unrestrained unless you are accounting for air resistance and then your vertical vector will be affected by gravity based on time...

 

[clintau]

No air resistance. The x component is 5 and y is 19.6. I don't know how to write the equation of the line. would it just be x=5.0? It doesn't seem that easy considering entire motion is parbolic? Would y=19.6 be correct for the motion in the y direction?

 

[Kurdt]

Have you heard of the kinematic equations for constant acceleration? You will need to use those.

 

[clintau]

no what are they?

 

[clintau]

do you mean:
xf=xi+vit+1/2at^2
and vf=vi+at

 

[Kurdt]

Here is a short list.

https://www.physicsforums.com/showpost.php?p=905663&postcount=2

 

[shamrock5585]

i believe you meant constant velocity... the velocity in the x direction is constant so you integrate with respect to time to get the position in terms of time... and then you have your velocity in the vertical direction which isn't constant because it is getting smaller due to gravity so you have to put that in and find WHEN the vertical velocity is zero so that you can integrate the entire vertical velocity equation to get position based in terms of time and plug in that time you just obtained to get max height...

 

[Kurdt]

You can use the constant acceleration equations with acceleration set at 0 and they reduce to those for constant speed.

 

[shamrock5585]

yes but if you start with 0 as your acceleration and integrate to get velocity what do you get?

 

[clintau]

xf=xi+vit+at2

xi=0
vi=5
a=?
t=4sec (or 2*tymax)
Xf=0+(5*4)+8a

yes no...I still don't know where this gets me toward writing an equation for the motion.

 

[clintau]

In review of the post it appears that I didn't know the RELEVANT EQUATIONS. I did. I have found ymax to be 19.6 and tymax to be 2.0

 

[Kurdt]

That is the equation of motion in the x direction. Acceleration is 0 by the way if you're ignoring air resistance.

 

[clintau]

if you have to use those equations in part a of the question why would my teacher ask for the equation of the motion in the x and y directions in part b?

 

[shamrock5585]

post what equations you have actually arrived at and maybe we can check what you got

 

[clintau]

Ymax=Yi+ViyT-4.9t^2
Ymax=19.6t-4.9t^2

Vf=Vi+at @ymax Vf=0 a=9.8(down)
t=2.0=Tymax

Ymax=0+19.6(2)-4.9(4)
Ymax=19.6

Correct

 

[shamrock5585]

looks good to me, that's what i got too... are you looking for anything else?

 

[clintau]

he asks for the equation of vertical motion y(t) and horizontal motion x(t)

I don't know how to express that

 

[Kurdt]

Those are the equations of motion. I don't know why your teacher has asked for them but there we go. Its as simple as that.

 

[clintau]

thank your help