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Objects thrown under free fall (1 Viewer)

rawker

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Hey all,
I got this question today:

16. Two objects are thrown off a cliff. One is thrown directly up, the other directly down. The ball thrown upward hits the ground
a. at the same speed as the one thrown downward
b. faster than the one thrown downward
c. slower than the one thrown downward

What do you's think?

Thanks
 

James747

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Is this suppse to be a trick question. The answer seems obviously to be c...
 

rawker

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I don't know. It can be all the answers, but I did choose c and I was wrong.
 

jyu

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rawker said:
Hey all,
I got this question today:

16. Two objects are thrown off a cliff. One is thrown directly up, the other directly down. The ball thrown upward hits the ground
a. at the same speed as the one thrown downward
b. faster than the one thrown downward
c. slower than the one thrown downward

What do you's think?

Thanks
Choice a.

The one thrown upward has the same speed downward when it passes through the point of projection.
 

xiao1985

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it's a

using equation
v2 = u2 + 2as

u is the same magnititude (but opposite in sign) for both scenarios, a and s are both the same for both scenarios... hence 2 v's equal to each other...
 

Bank$

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shouldnt it be "b" ? because the one thrown up will accelerate at 9.8 for a greater distance than the one thrown down. The only other answer that makes sense is "a" BUT ONLY WHEN CONSIDERING THAT BOTH REACH TERMINAL VELOCITY.


Hey Xiao in "V^2=U^2 + 2as"

isnt the S different since the one thrown up has a bigger S and hence V cannot be equal ????
 
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ianc

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no - remember S is displacement, not distance travelled. So the displacement is equal when both balls have hit the ground because they were both launched from the same position.

(i.e. S for both of the equations would be the distance from the ground to the top of the cliff)

edit: also in this question, we are ignoring air resistance and all other forces, so there is no terminal velocity to worry about (terminal velocity would occur when the force due to air resistance was equal in magnitude to that of gravity)
 
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Bank$

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oh rite yeah S is displacement

but when the one that is fired up falls bak down to level with the cliff should it have an initial velocity unike the other one that started with zero velocity at the same level ???? and therefore hit the ground faster as can be seen in v^2 = u^2 + 2as
 

Forbidden.

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All objects in free-fall undergo constant acceleration, when there is no/constant acceleration there is no change in speed therefore both objects fall with the same speed, need I explain my theory with calculus ?

a = dv/dt

Let v = Velocity of object and t the time of flight (or fall) ...

:burn:
 

Not-That-Bright

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Think about it... when you throw a ball up, it comes back at you at the same speed as you initially threw it, correct? The speed of the ball comming back down will be going as fast as it was going up - at the point of initial departure. Therefore throwing a ball up or down (in a situation where there is no resistance mind you) with the same initial speed will mean they both land with the same speed.
 
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Bank$

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"Not-That-Bright" u r VERY BRIGHT ur simple explanation made perfect sense to me :)
 

twilight1412

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answer is b

remember you are throwing off a cliff, if you throw it up it will be under the affect of gravitation so if you think about it carefully

v = u + at

because its under a for a longer amount of time it will accelerate more and therefore will have a greater velocity
 

Not-That-Bright

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because its under a for a longer amount of time it will accelerate more and therefore will have a greater velocity
No because you have to take account the u of the object being thrown down. The object being thrown up will only accelerate to a speed of the u of the object being thrown down at the same point.

It is in the air longer to accelerate more however all that the extra time makes up for is to catch up to the other object's u.
 

twilight1412

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i guess the main decision is how we define throw because to throw downwards could also be considered dropped in some cases
 

xiao1985

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twilight1412 said:
i guess the main decision is how we define throw because to throw downwards could also be considered dropped in some cases
rarely... and if no air resistance is considered, b is incorrect...
 

Mat_Aussie

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it all depends on the height of the cliff.

lets break it down:

if the cliff is really high :
The ball being thrown downwards (has force of throw to accelerate) may reach terminal veolcity faster than the ball being thrown upwards as it relies solely on gravity for its accleration. But each ball can only reach a certain speed (terminal velocity) therefore resulting in both balls impacting at the same velocity. if that all makes sense.


if the cliff is reatively low: ( balls unable to reach terminal velocity)
you would think that because the ball thrown up would be faster because gravity will act longer thus resulting in a higher velocity at impact than the ball thrown down.

however this isnt necessarily the case, since the balls are being thrown and not dropped. the ball thrown downwards is given a certain velocity at t = 0. and then has x amount of time to accelerate due to gravity. whereas the ball thrown upwards loses its given velocity as it reaches the apex/top of the throw. this ball now how has a higher position to start its acceleration, however the extra velocity achieved with the extra height may not be greater than the velocity given to the downwards ball at t = 0.

therefore you cannot determine the answer without knowing the hight of the cliff.

i think that it all makes sense...
 

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