Projectile Motion Confusion!

[axp107]

Is the velocity at the top of its trajectory ZERO? EK says it is! But isn't there always a horizontal velocity thats constant.. how can V at the top be 0.

Look at this problem

1) If an antelope is running at a speed of 10m/s and can maintain that horizontal velocity when it jumps, how high must it jump to clear a horizontal distance of 20 m?

I'm getting confused by the term "Velocity". Is it referring to horizontal or vertical velocity.

What about this one?

2) A projectile is launched with an angle of 30 degrees and an initial speed of 100 m/s. What is the velocity of the projectile at its highest point?

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[Vanguard23]

The second one is the cosine of 100m/s at 30 degress, since it is only the horizontal you are considering. Should be 86m/s.

 

[evoviiigsr]

The vertical COMPONENT (vector) of the velocity is zero at the moment when the projectile is at its highest point. Horizontal velocity, as you said, remains constant throughout.

For the projectile question, you know that the total velocity is 100 m/s. You also know the angle from which it was launched. At the highest point, there is no vertical velocity component, so you can simply find the horizontal component and that will be the total velocity as well. So cos 30 * 100= 87 m/s is the horizontal and TOTAL velocity at the object's highest point.

 

[axp107]

Ok, I get that part

Now can anyone help me with the deer question?

 

[shaqkillz451]

The Vertical Velocity at the top is zero, not the total velocity...like u said, there's always a constant horizontal velocity since no acceleration in that direction (assuming no air resistance)...the only time total velocity is zero if something is thrown straight up...

For the first question, time in air must be 2 seconds (x=vt, solve for t)...means that vertical velocity must be such that deer takes 1 second to get to the top...therefore, initial vertical velocity must be 10 m/s.....to find out how high the deer will go at this initial vertical velocity, y=((Vi + Vf)/2)t....solve for y given that Vi=10 and Vf=0...height comes out to be 5 meters......plzzz correct me if i'm wrong

 

[axp107]

The Vertical Velocity at the top is zero, not the total velocity...like u said, there's always a constant horizontal velocity since no acceleration in that direction (assuming no air resistance)...the only time total velocity is zero if something is thrown straight up...

For the first question, time in air must be 2 seconds (x=vt, solve for t)...means that vertical velocity must be such that deer takes 1 second to get to the top...therefore, initial vertical velocity must be 10 m/s.....to find out how high the deer will go at this initial vertical velocity, y=((Vi + Vf)/2)t....solve for y given that Vi=10 and Vf=0...height comes out to be 5 meters......plzzz correct me if i'm wrong

Thats the right answer...

1) But I don't get why you said the INITIAL vertical velocity is 10 m/s... Doesn't it say the horizontal velocity is 10m/s???

Is it from V=V0 + at.. nvm I see... how'd you figure that out though.. so easily.. and shortcut?

2) Also can't you do x= x0 +V0t +1/2 a*t^2 .... where V0 = 10m/s (vertical) to find x, the height where t=1?

Thats giving me an answer of 15 🙁 ... why'd u use average velocity formula 🙁

 

[shaqkillz451]

do u understand that time in the air has to be 2 seconds?....If you do, then it should also make sense that the time it takes to reach max height (the point where vertical velocity=0) should be 1 second (1 second to go up, 1 second to come back down).

if g=10, and we want the final vertical velocity to be zero after 1 second, what should be the initial vertical velocity? It should be 10 right? If that doesnt make sense, note what is known: t=2, a=-10, Vf=0, and we want Vi....so Vf=Vi +at, solve for Vi, it should be 10m/s.

As long as you understand that time in the air must be 2 seconds.... everything else is easier

 

[shaqkillz451]

u can use x=xo +vot +1/2at^2, just remember that a is -10, so it should be

x= xo + vot + .5at^2 OR
x= 0 + 10(1) + .5(-10)(1) = 5 meters

all these equations can be derived from each other, so use whichever one is more convenient

 

[axp107]

u can use x=xo +vot +1/2at^2, just remember that a is -10, so it should be

x= xo + vot + .5at^2 OR
x= 0 + 10(1) + .5(-10)(1) = 5 meters

all these equations can be derived from each other, so use whichever one is more convenient

Why'd you choose g to be negative

Is it always: going up g is -ive.. coming down, g is +ive? b/c A and V are in opposite directions...

I guess I'm having more trouble with projectile problems.. b/c I always make g +ive no matter what.. I can't just blindly do that can I? Aah, no wonder I get falling down problems right in general... and mess up on Projectiles

I still don't see why you chose to use the Average Velocity formula. That way, you bypass the whole gravity sign confusion. Any logic behind why you automatically used that formula?

 

[Richter915]

do u understand that time in the air has to be 2 seconds?....If you do, then it should also make sense that the time it takes to reach max height (the point where vertical velocity=0) should be 1 second (1 second to go up, 1 second to come back down).

if g=10, and we want the final vertical velocity to be zero after 1 second, what should be the initial vertical velocity? It should be 10 right? If that doesnt make sense, note what is known: t=2, a=-10, Vf=0, and we want Vi....so Vf=Vi +at, solve for Vi, it should be 10m/s.

As long as you understand that time in the air must be 2 seconds.... everything else is easier

are you sure that's right? When you plug the numbers into the equation you posted you do not get 10 for Vi. I think you meant t=1 over there. I think the crucial aspect to this question is knowing that time of max height and V=0 is at t=1.

 

[axp107]

are you sure that's right? When you plug the numbers into the equation you posted you do not get 10 for Vi. I think you meant t=1 over there. I think the crucial aspect to this question is knowing that time of max height and V=0 is at t=1.

Yeah.. he did use t = 1

 

[Richter915]

Why'd you choose g to be negative

Is it always: going up g is -ive.. coming down, g is +ive? b/c A and V are in opposite directions...

I guess I'm having more trouble with projectile problems.. b/c I always make g +ive no matter what.. I can't just blindly do that can I? Aah, no wonder I get falling down problems right in general... and mess up on Projectiles

I still don't see why you chose to use the Average Velocity formula. That way, you bypass the whole gravity sign confusion. Any logic behind why you automatically used that formula?

since acceleration is a vector, you need to identify the direction. When something is shot upwards, gravity pulls it DOWN, right? Using cartesian coordinates (I think), you can say that anything working in the down direction is negative, anything up is positive. You just need to stay consistent.

If you chose to make g a positive value, then you would have to make Vi a negative value because Vi is in the opposite direction as g. Hope this makes sense.

 

[cheezer]

Thats the right answer...

1) But I don't get why you said the INITIAL vertical velocity is 10 m/s... Doesn't it say the horizontal velocity is 10m/s???

Is it from V=V0 + at.. nvm I see... how'd you figure that out though.. so easily.. and shortcut?

2) Also can't you do x= x0 +V0t +1/2 a*t^2 .... where V0 = 10m/s (vertical) to find x, the height where t=1?

Thats giving me an answer of 15 🙁 ... why'd u use average velocity formula 🙁

First off when the other posters said vertical velocity is 0, they mean the net y component of velocity is 0. There's still a velocity upwards, but at one point, when gravity is busy overtaking the deers thrust upward, the net velocity is 0.

1. He figured it out because the problem gives only the horizontal velocities. The only thing that can relate the horizontal with the vertical velocities (since they are independant) is the time. Thus he calculated the total time it took to clear the field horizontally and reasoned that half that time would be the peak, or where the net y velocity would equal 0.

At 1 second, he found the initial y velocity, yes, you were correct, using: v=vo-gt. Calling v=0, and calling g=10m/s, he found that v0 in the y velocity was 10m/s.

He subsequently calculated the peak height using y=((Vi + Vf)/2)t

You can also use y=y0+v0t- 1/2 gt^2
y0=0
v0=10
t=1
g=10

y=0+10- 5=5

to answer your question about the "average velocity forumla" y=((Vi + Vf)/2)t...the use is correct
because it basically says, the average velocity between the y=0 and y=5
multiplied by the time it takes to get from y=0 and y=5 will give you y=5. god i hope didnt explain too poorly
*g is not used to find the final answer because you already know vi and vf. remember you calculated vi using the g. vf takes the g in consideration also, although you know conceptually that it'll be 0

 

[shaqkillz451]

Good explaination cheezer!

 

[cheezer]

like richter says, your confusion with the signs has everything to do with your coordinate system. by saying gravity is positive, you're essentially calling the direction downward positive, and the direction upward negative.

by doing that you call v0 in the y direction -10m/s. that's okay because the thrust is upward and upward is "negative" in this coordinate system. then, by solving for y, you get -5 m. that too is also okay because upward in this coordinate system is negative. and you're essentially calling 5 m up the negative y axis the peak

 

[cheezer]

Good explaination cheezer!

it was a collective effort...

 

[Richter915]

it was a collective effort...

if only the actual mcat could be a group effort too.... 🙁

 

[axp107]

Thanks for the explanation.

Basically, I'm going to do it this way.. seems consistent:

- Make Velocity positive ALWAYS... change the sign of gravity accordingly.

 

[Richter915]

Thanks for the explanation.

Basically, I'm going to do it this way.. seems consistent:

- Make Velocity positive ALWAYS... change the sign of gravity accordingly.

sounds like a plan to me...just keep it consistent throughout your problem and you should be good.

funny story about this...with the equation Vf^2 = Vo^2 + 2a(deltax)...my prof ALWAYS wrote it as Vf^2 = Vo^2 - 2g(deltax)...so come test time, he throws in a problem where you have to make acceleration positive and pretty much everyone got it wrong lol.

 

[DrBowtie]

Thanks for the explanation.

Basically, I'm going to do it this way.. seems consistent:

- Make Velocity positive ALWAYS... change the sign of gravity accordingly.

Just make down negative and up positive. Thats the simplest.

 

[cheezer]

Thanks for the explanation.

Basically, I'm going to do it this way.. seems consistent:

- Make Velocity positive ALWAYS... change the sign of gravity accordingly.

well just to clarify, you just need to be aware of the directions.

so first picture your coordinate system, you call up +y, down -y, right +x, left -x. IF the deer hops up and to the right, only then is velocity in the y and x directions positive. gravity in this system is always going to be -y, with no x component.

 

[axp107]

well just to clarify, you just need to be aware of the directions.

so first picture your coordinate system, you call up +y, down -y, right +x, left -x. IF the deer hops up and to the right, only then is velocity in the y and x directions positive. gravity in this system is always going to be -y, with no x component.

yup, thanks for confirming.
Thanks for the help guys.. man I took physics so long ago.. getting a little rusty.. it's all good though. I've got until august

took the kaplan diagnostic today.. thought PS was brutal.. bio was ok.. verbal went fine (hopefully) =D

 

[axp107]

Just for clarification:

Are these two equivalent?

1) making up +ive and down -ive (gravity and velocity)

2) ALWAYS keeping velocity +ive, but making g -ive when going up.. and g+ive when going down.

The reason I'm asking is that, I remember to put the sign on gravity.. but I tend to forget to put the sign on Velocity. Will my method (number 2) give correct answers?