Dennis and Carmen problem with my trial solution , is it right?

  • Thread starter Thread starter nareeman
  • Start date Start date
  • Tags Tags
    trial
AI Thread Summary
Dennis and Carmen throw basketballs in opposite directions from a cliff, and the discussion revolves around determining which ball hits the ground faster. Both balls have the same initial speed, but Dennis's ball takes longer to return to the initial position before falling. The key point is that when both balls hit the ground, they will have the same final speed due to the conservation of energy principle, despite the different paths taken. The confusion arises from the mathematical treatment of initial velocities, but the conclusion remains that both balls will hit the ground at the same speed. Understanding this concept is crucial for correctly answering related exam questions.
nareeman
Messages
6
Reaction score
0

Homework Statement



Dennis and Carmen are standing on the edge of a cliff. Dennis throws a basketball vertically upward, and at the same time Carmen throws a basketball vertically downward with the same initial speed. You are standing below the cliff observing this strange behavior. Whose ball is moving fastest when it hits the ground?

Homework Equations



v^2 = v0^2 -2gΔy


The Attempt at a Solution


in the same speed cause when Dennis basketball back to the initial position "`the throwing position" it will back with the same initial speed with different direction "the same direction as Carmen" so it's the same but it'll take Dennis basketball more time
Dennis
v0 is +
v^2 =+v0^2 -2gΔy
Carmen
v0 is -
v^2 =-v0^2 -2gΔy
-----
g =-9.8 for both
Δy is the same for both


so
Dennis v^2 =+v0^2
Carmen v^2 =-v0^2
add them
v^2+v^2 =0

v^2 =-v^2

please tell me if that's right if not how it can be solve
thank you:smile:
 
Physics news on Phys.org
nareeman said:

Homework Statement



Dennis and Carmen are standing on the edge of a cliff. Dennis throws a basketball vertically upward, and at the same time Carmen throws a basketball vertically downward with the same initial speed. You are standing below the cliff observing this strange behavior. Whose ball is moving fastest when it hits the ground?

Homework Equations



v^2 = v0^2 -2gΔy


The Attempt at a Solution


in the same speed cause when Dennis basketball back to the initial position "`the throwing position" it will back with the same initial speed with different direction "the same direction as Carmen" so it's the same but it'll take Dennis basketball more time
Your logic is correct :approve:

Dennis
v0 is +
v^2 =+v0^2 -2gΔy
Carmen
v0 is -
v^2 =-v0^2 -2gΔy
Careful, the formula wants you to square the initial velocity. Carmen's initial velocity is -v0, and squared is (-v0)(-v0) = +v02.

As you can see, the initial velocity being negative does not change the result.

-----
g =-9.8 for both
Δy is the same for both


so
Dennis v^2 =+v0^2
Carmen v^2 =-v0^2
add them
v^2+v^2 =0

v^2 =-v^2
If you consider that last line of math it represents an impossible situation, since you can't have any real number squared that turns out negative. The glitch can be traced back to squaring the initial velocity as I pointed out above.

You could also have used a conservation of energy approach (if you've covered that yet in your course).
 
  • Like
Likes 1 person
thank you ,we have not cover it yet
so if I'm going to answer for that question in the exam i just write this formula "v^2 = v0^2 -2gΔy" and say
everything is the same for both cases " v0^2 -2gΔy"
then the final speed is the same
only or should i add something else ?
 
nareeman said:
thank you ,we have not cover it yet
so if I'm going to answer for that question in the exam i just write this formula "v^2 = v0^2 -2gΔy" and say
everything is the same for both cases " v0^2 -2gΔy"
then the final speed is the same
only or should i add something else ?

Forget the math for a minute and just think about it this way. If the ball thrown upward has an initial velocity X, then when it gets back to the same point on its way down, what is its velocity at that point?
 
nareeman said:
thank you ,we have not cover it yet
so if I'm going to answer for that question in the exam i just write this formula "v^2 = v0^2 -2gΔy" and say
everything is the same for both cases " v0^2 -2gΔy"
then the final speed is the same
only or should i add something else ?

That's all you really need :smile:
 
phinds said:
Forget the math for a minute and just think about it this way. If the ball thrown upward has an initial velocity X, then when it gets back to the same point on its way down, what is its velocity at that point?

the same initial velocity that was it thrown with
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

Similar threads

Replies
11
Views
4K
Replies
5
Views
2K
Replies
23
Views
3K
Replies
5
Views
1K
Replies
5
Views
2K
Replies
9
Views
4K
Back
Top