projectile motion question

[Mae16]

can anyone help with a quick question?

if two identical balls are launched from a cliff, and one is launched with just a horizontal velocity (let's say 50m/s) and the other is also launched at 50m/s but at some angle (so it has a vertical velocity), why do they hit the ground with the same velocity?

I understand from a conservation of energy standpoint - they both start at the same potential and kinetic energy -- so they must end up with the same KE, and thus end with the same velocities as each other.

but from an intuitive standpoint, it seems that because one is launched up into the air, it has more time for the acceleration due to gravity to act on it. let's say that it starts freefalling from 100m; and the horizontal ball only starts freefall from 50m -- wouldn't then the first ball be subject to g over a longer timeframe and thus hit the ground with a faster velocity?

what am I missing here?

thanks!

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[Bernoull]

can anyone help with a quick question?

if two identical balls are launched from a cliff, and one is launched with just a horizontal velocity (let's say 50m/s) and the other is also launched at 50m/s but at some angle (so it has a vertical velocity), why do they hit the ground with the same velocity?

I understand from a conservation of energy standpoint - they both start at the same potential and kinetic energy -- so they must end up with the same KE, and thus end with the same velocities as each other.

but from an intuitive standpoint, it seems that because one is launched up into the air, it has more time for the acceleration due to gravity to act on it. let's say that it starts freefalling from 100m; and the horizontal ball only starts freefall from 50m -- wouldn't then the first ball be subject to g over a longer timeframe and thus hit the ground with a faster velocity?

what am I missing here?

thanks!

2 words .. Terminal Velocity!!

The implicit assumption in your statement is that the cliff is high enough for the ball(s) to attain terminal velocity. Another assumption is the presence of air resistance.

Basically after objects are dropped, the accelerate by g, and force of air resistance builds up with increased velocity to a point where it exactly counters gravitational force. At this point, u have dynamic equilibrium = no acceleration therefore IDENTICAL balls will have equal final/terminal velocity.

The fact that force of air resistance increases with increased velocity is very intuitive, stick ur hand outside ur car while accelerating...

 

[Dr Gerrard]

Well still, wouldn't one of the balls reach terminal velocity before the other, thus falling to the ground faster?

If you start out at a negative velocity, while the other one starts out at a velocity of 0, the one with the negative velocity would take longer to reach a velocity of 10 than the one with a 0 velocity, given the same force. Correct?

 

[Bernoull]

Well still, wouldn't one of the balls reach terminal velocity before the other, thus falling to the ground faster?

If you start out at a negative velocity, while the other one starts out at a velocity of 0, the one with the negative velocity would take longer to reach a velocity of 10 than the one with a 0 velocity, given the same force. Correct?

Everything you said is true EXCEPT ur implication that reaching terminal velocity (TV) sooner equals greater final velocity. By definition, terminal velocity is the final/maximum velocity (for freefalls).

Think of it this way, from F=ma, to change velocity, u must have a net force. At TV, there's a dynamic equilibrium thus no net force which means no acceleration. Balls or anything for that matter, won't freefall any faster bcos there's no net force..

 

[Dr Gerrard]

What I meant was not that the terminal velocity would be greater, but that the terminal velocity would be reached earlier in the ball that starts out at 50m/s only horizontal velocity, with 0 vertical velocity, than the ball that starts out with a 50 m/s upwards velocity.

Since it would take longer for the ball moving upwards to reach terminal velocity, it would talk longer for it to hit the ground.

 

[matth87]

Are there any more variables in the problem?

 

[matth87]

I see responses that deal with terminal velocity and forces... This is a simple kinematics problem dealing with 5 variables. a, t, vf, vo, d.

You dont have to use forces...

Where did you get this question from?

 

[Bernoull]

Always break your velocities into their components when doing problems like this.

The Ball 2 has velocity in the X of 30m/s and a velocity in the Y of 0m/s

The Ball 1 has a velocity in the Y of Vo Sin(theta)m/s. The final velocity of this ball will be greater since it has a downward velocity to begin with or you can think of it like "its being dropped from a higher position". Acceleration will have more time to act on it to speed it up.

Maybe you confused the question with that they would hit at the same time possibly?

Let me know if you have any questions

**********EDIT, is the answer looking for the velocity in the Y direction or the net velocity? does it give angle theta?

OP ALREADY states final velocities of both balls are EQUAL, ur answer ignores this fact...

 

[DOC209]

can anyone help with a quick question?

if two identical balls are launched from a cliff, and one is launched with just a horizontal velocity (let's say 50m/s) and the other is also launched at 50m/s but at some angle (so it has a vertical velocity), why do they hit the ground with the same velocity?

I understand from a conservation of energy standpoint - they both start at the same potential and kinetic energy -- so they must end up with the same KE, and thus end with the same velocities as each other.

but from an intuitive standpoint, it seems that because one is launched up into the air, it has more time for the acceleration due to gravity to act on it. let's say that it starts freefalling from 100m; and the horizontal ball only starts freefall from 50m -- wouldn't then the first ball be subject to g over a longer timeframe and thus hit the ground with a faster velocity?

what am I missing here?

thanks!

the way i see it the ball launched horizontally will eventually start falling and have the same exact force acted on it as the ball launched at an angle...thus by the time they hit the ground both should have the same same V final...

 

[Bernoull]

the way i see it the ball launched horizontally will eventually start falling and have the same exact force acted on it as the ball launched at an angle...thus by the time they hit the ground both should have the same same V final...

Are you accounting for air resistance or assuming it's negligible?? If it's negligible the ball shot at an angle MUST have a higher final velocity bcos it's flight time is longer (Vf=at; Vo is zero).

The easiest way to look at this problem is to realize that vectors act independently so completely ignore the horizontal velocities of both balls (they don't affect flight time). At release, Ball 1 has 0m/s vertical and Ball 2 has a positive upward vertical velocity. To simply u can pretend that Ball 2 has no upward vertical velocity but it's simply dropped from a higher height (basically KE is converted to GPE). With 2 balls dropped from different heights, the higher the height, the faster the final velocity. The ONLY time this isn't the case is when air resistance is present and the heights are sufficiently high for terminal velocities to be achieved and then final velocities are equal...

 

[Fort]

OP, I think you're better off looking at it from an energy standpoint. I understand you want to see the logic in it from a different angle, but lots of motion problems are much more easily solved using conservation of energy. In fact, I usually try that first before trying to tackle a problem using any of the 5 kinematic equations.

 

[deleted252565]

I think looking at either kinematic or energy standpoint doesn't give the same final velocity. From energy standpoint, ball 1 starts with v_i = 0 (which has PE, but no KE) whereas ball 2 starts with v_i # 0 (thus has both PE + KE) ==> meaning their final v shouldn't be the same, and ball 2 should have higher final v.

From kinematic standpoint, ball 2 has initial vertical velocity, thus after it travels upward until v_vertical = 0, it falls back down and obtain the same velocity at the position where it initially left, then continues falling downward. Thus, ball 2 basically reaches the bottom with diff. velocity, or higher velocity.

My reasoning ignores air resistance.

 

[Bernoull]

I think looking at either kinematic or energy standpoint doesn't give the same final velocity. From energy standpoint, ball 1 starts with v_i = 0 (which has PE, but no KE) whereas ball 2 starts with v_i # 0 (thus has both PE + KE) ==> meaning their final v shouldn't be the same, and ball 2 should have higher final v.

From kinematic standpoint, ball 2 has initial vertical velocity, thus after it travels upward until v_vertical = 0, it falls back down and obtain the same velocity at the position where it initially left, then continues falling downward. Thus, ball 2 basically reaches the bottom with diff. velocity, or higher velocity.

My reasoning ignores air resistance.

OP states that both balls at launched at 50/ms but at different directions. Energy being a scalar quantity does not involve direction so both balls have EQUAL KE and PE at launch.

This is a ridiculously simple problem, OP needs to state whether problem assumes no air resistance or not!! If air resistance is at play, mechanical energy is NOT conserved (u may still get away with conservation of energy approach since balls are identical) but kinematics still works..

OP is there air resistance or not?

 

[Mae16]

thanks for all the responses.

the question is from tbr physics part I, passage 5 (#35), and it does telll us to ignore air resistance.

I totally get it from a conservation of energy perspective -- KE & PE for both balls at launch is identical; therefore KE & PE for both balls at end must be identical; and therefore velocities of both balls at end must be identical.

but I still don't get it from a kinematics standpoint. if one ball is launched higher up into the air, it still seems like g should act on it for longer and its final velocity should be greater.

 

[matth87]

thanks for all the responses.

the question is from tbr physics part I, passage 5 (#35), and it does telll us to ignore air resistance.

I totally get it from a conservation of energy perspective -- KE & PE for both balls at launch is identical; therefore KE & PE for both balls at end must be identical; and therefore velocities of both balls at end must be identical.

but I still don't get it from a kinematics standpoint. if one ball is launched higher up into the air, it still seems like g should act on it for longer and its final velocity should be greater.

It might be because you have consider x=0 point and make a coordinate system so the ball actually has a initial negative velocity. Just speculation, this problem is keeping me up at night though lol. Im gonna make up some numbers and plug them in to see if it works.

 

[Bernoull]

This is a classic example of why energy approach is almost always better than vector-based quantities like velocity. Qualitatively,

[GVIDEO][/GVIDEO]​

Since air resistance is negligible, horizontal velocity is undiminished throughout. Ball A begins wit pure horizontal v an descends from a lower max height hence final vertical velocity is less than Ball B's. Now Ball B has a smaller horizontal velocity (50*cosQ) but has a greater final vertical velocity bcos max GPE is greater than Ball A. Now velocity being a vector, u add da x&y components for both balls and I think they should be equal.. I'll try to state this quantitatively.

 

[ssiding]

So lets pick some arbitrary numbers and then make sure our coordinate system is in place.

I choose the acceleration due to gravity as 10m/s2 and the height of the cliff to be 5m, and down as the positive direction!!

Ball 1
Balll 1 only has initial velocity of 50m/s in the X direction so lets figure out its over all final velocity taking into account the final X and Y velocities

Final Y velocity= using the formula Vf^2 = Vo^2 + 2ad ... since Vo in the Y direction is zero we get Vf^2 = 2ad... Vf^2= 2(+10m/s2)(+5m).. solve for Vf is 10m/s

Final X velocity= the X velocity remain the same throughout the flight... thus 50m/s

Final over all velocity of Ball 1= using pathagorous we get the overall final velocity as around 51m/s


Ball 2
Remember that we set the down direction as the positive
Ball 2 is launch at an angle of 30 degrees above the horizontal at 50m/s thus it has an initial velocity is the X and Y direction...so lets solve for these seperately as we did above for Ball 1.

Final Y velocity = using the formula Vf^2 = Vo^2 + 2ad.. remember that since its launch at an angle the Y veocity will be sin(angle) * hypotneuse which is 50m/s... and also remember that we will have an negative initial velocity in the Y direction for ball 2.. thus Vf^2= Vo^2 + 2ad... Vf^2= (sin(30)*(-50m/s))^2 + 2(+10m/s2)(+5m)... solve for Vf in the Y direction and we get around 27m/s.

Final X velocity = using using the formula Vf^2 = Vo^2 + 2ad...Vf^2= Vo^2 + 2ad... Vf^2= (cos(30)*(50m/s))^2 + 2(+10m/s2)(+5m)...solve for Vf and we get 45m/s

Final overall velocity for Ball 2
Again using pathagorous.. we get our overall velocity for ball 2 as being aroung 52m/s which is pretty close... not exactly 51 as for ball 1 but that might just be due to rounding.

 

[iPodtosis]

ssiding... that seems like a lot of math for MCAT physics

can anyone write out the way to do it in Energy Conservation?
Thanks

 

[Doodl3s]

@ iPodtosis: Its LITERALLY mgh + (1/2)mv^2 = (1/2)mv^2
where h is the height of the cliff, and the first v is the velocity of the ball in WHATEVER direction you shoot it, no matter the angle the final v will be the velocity it hits the ground with!
Its that simple. The cliff height adds its potential energy to kinetic you gave it at the beginning. It can be explained intuitively as it has been many times above. If you shoot the ball straight down you have that start velocity plus watever it accelerates to. If you shoot it straight up, it will pass its point of origin at the same speed you shot it up, but instead will be going straight down at the speed (only matters if time is involved, otherwise its exactly the same) if you shoot horizontally your gonna have a horizontal and vertical vector that will come together to an overall velocity of whether you shot it straight up or down...

 

[wrathofgod64]

Doodl3s is right. The OP has been only taking into account vertical velocity. You have to account for the horizontal component to velocity that exists for the ball launched horizontally. So while the ball launched straight up will have a greater vertical component to its velocity, and will be traveling faster in the downward direction, it will still have the same velocity as the horizontal ball, once u add the latter's components together.

 

[ssiding]

yea I definately agree that using kinematics would require too much math, and from doodles post COM is alot easier.

I guess thats a good trick to use...if the math looks like it will take too long then there must be a better way to do the problem... from what i've heard its should not take more than 45s-60sec to work out the math on the mcat any more then you've choosen the wrong concept

sorry if i'm rambling..i've been up all night