This should be an easy question, but I think I'm tricking myself.

[kajasu88]

Two identical balls are launched with the same initial speed. The first ball is launched directly upwards and the second ball is launched directly horizontally. Do the balls hit the ground at the same time? Do the balls hit the ground with the same total speed?

 

[rock.freak667]

Use the projectile motion kinematic equations for each case and compare.

 

[InTuoVultu]

keep in mind where the ground is. If you're starting at ground level and shoot horizontally, you're never in the air so time of flight is zero.
If both of them start at some height h, then the horizontal one lands first. Your airtime is only determined by the y position and velocity.
Both of the balls started with the same energy, so neglecting air resistance, both of them should have the same energy when they hit the ground and hence the same speed, albeit in different directions. The vertical one will be going straight down and the horizontal one will be at some slant.

If you do take into account air resistance, the vertical one will have less energy. I think. At least that's what my physics gut tells me. You'd have to figure out some slick way of calculating that if you want to because I know that at least for 2-D projectile motion with drag the equations are unsolvable. At least with normal techniques.

 

[Zaphys]

I know that at least for 2-D projectile motion with drag the equations are unsolvable. At least with normal techniques.

What do you mean with that? Of course they're solvable, they happen to be separable diferential ecuations of first order for velocity as function of time. The derivation of the results, i.e. final time or speed, may be a little more tricky but with a normal calculator you´ll be able to get them.

I´m not sure if you referred to that but hope it helps :)

Zap

 

[Redbelly98]

If both of them start at some height h, then *** lands first.

Please don't give out answers in the future. Our practice is to give help in the form of hints, and let the OP's brain do as much of the work as possible.

If you do take into account air resistance ...

Since this is an introductory physics problem, it's safe to assume that air resistance is to be neglected

 

[Red_CCF]

keep in mind where the ground is. If you're starting at ground level and shoot horizontally, you're never in the air so time of flight is zero.
If both of them start at some height h, then the horizontal one lands first. Your airtime is only determined by the y position and velocity.
Both of the balls started with the same energy, so neglecting air resistance, both of them should have the same energy when they hit the ground and hence the same speed, albeit in different directions. The vertical one will be going straight down and the horizontal one will be at some slant.

If you do take into account air resistance, the vertical one will have less energy. I think. At least that's what my physics gut tells me. You'd have to figure out some slick way of calculating that if you want to because I know that at least for 2-D projectile motion with drag the equations are unsolvable. At least with normal techniques.

forgive me if I'm wrong but I don't think that's correct

 

[Redbelly98]

Since the number of responses here is growing, I'll just point out that rock.freak667 has provided the only one that is directly useful to kajasu88:

Use the projectile motion kinematic equations for each case and compare.

p.s. welcome to Physics Forums kajasu88

 

[kajasu88]

wow thanks for all the help!

this is my first physics class and all the theories are starting to get mixed up in my head.

 

[Red_CCF]

wow thanks for all the help!

this is my first physics class and all the theories are starting to get mixed up in my head.

Is your teacher asking you to find a value or just state one's velocity/time of impact relative to the other?

 

[kajasu88]

No values, just relative to each ball.
ie:
Do the balls hit the ground at the same time?
y/n. which one would hit first? horizontal ball/vertical ball
Do the balls hit the ground with the same total speed? y/n

 

[Red_CCF]

No values, just relative to each ball.
ie:
Do the balls hit the ground at the same time?
y/n. which one would hit first? horizontal ball/vertical ball
Do the balls hit the ground with the same total speed? y/n

What do you think and why? I'm not allowed to tell you what I think is the right answer

 

[kajasu88]

same speed, different times
the one thrown vertically will take longer, i don't really know why, I would just imagine that it does. and same speed because they have the same initial velocity... ?

 

[berkeman]

same speed, different times
the one thrown vertically will take longer, i don't really know why, I would just imagine that it does.


and same speed because they have the same initial velocity... ?

Stop guessing. Use the suggestion from post #2 and stop trying to cheat. You are in school. Do the work.

 

[kajasu88]

I'm not guessing. I'm going off an experiment in lab when he put two balls on a track, one track was curvy and the other was straight. I think this is the same concept. The ball on the curvy track finished first b/c it gains PE as it goes up and down the curves. They also finished at the same speed.

 

[berkeman]

I'm not guessing. I'm going off an experiment in lab when he put two balls on a track, one track was curvy and the other was straight. I think this is the same concept. The ball on the curvy track finished first b/c it gains PE as it goes up and down the curves. They also finished at the same speed.

Wonderful if true. Post your work based on post #2 to the original post (OP) question.

 

[Redbelly98]

Another hint, besides the excellent one by rock.freak in post #2:

To answer the question about hitting the ground at the same time, look at the vertical component of motion. Ignore the horizontal motion for this question.

 

[Red_CCF]

For this problem you don't even need projectile motion. Just look at the energy of the verticle component of each one and you can determine the velocity of each one.

 

[Redbelly98]

That works only if the class has discussed kinetic energy. They may or may not have covered that yet.

 

[kajasu88]

we did do KE, but can u separate KE in components?

 

[Red_CCF]

we did do KE, but can u separate KE in components?

well don't think of it like components of vectors because energy is NEVER a vector. I would imagine it as the total energy equal to the energy of the verticle motion + energy of the horizontal motion. It's more like it's split with part of the energy going horizontal and part of it going towards verticle motion. You're only interested in the vertical part so you have to break it down.

 

[kajasu88]

okay so I did:
Ball 1 (vertical ball)
x: 1/2m(vf^2 - 0)
y: 1/2m(vf^2 - vi^2)

Ball 2 (horizontal ball)
x: 1/2m(vf^2 - vi^2)
y: 1/2m(vf^2 - 0)

solve for vf
pythagorean theorem
so the vf is the same.

 

[Red_CCF]

okay so I did:
Ball 1 (vertical ball)
x: 1/2m(vf^2 - 0)
y: 1/2m(vf^2 - vi^2)

Ball 2 (horizontal ball)
x: 1/2m(vf^2 - vi^2)
y: 1/2m(vf^2 - 0)

solve for vf
pythagorean theorem
so the vf is the same.

By using the pythagorean theorem you're essentially adding vectors using energy which is wrong. Try it this way, find an expression for the total energy of each ball initially (which should be the same) but with one of the balls and find the energy that it expends going horizontally and vertically. You should see that the energy that the two object expends vertically is different.

 

[kajasu88]

is the change in KE = 0?? since there is no external forces?

 

[kajasu88]

well...

in the y direction:
Ball one, v initial is w/e v initial is (no value)
Ball two v initial is zero.

 

[kajasu88]

and vice versa for the x direction

 

[Red_CCF]

well...

in the y direction:
Ball one, v initial is w/e v initial is (no value)
Ball two v initial is zero.

You're on the right track. I think it will help if you drew this stuff out and labelled it. With physics it's easier to do the question if you can see what is going on.

With the ball going straight up, we only have one component, the vertical, so all the energy must be in the vertical component. But this is not the case for other ball because some energy is expended in the horizontal component. Since the total energy of each of the ball is equal, then the energy of the vertical motion of each ball is different. We can ignore the horizontal portion of the 2nd ball because we don't care about it.