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1. Evening maths people, I've been looking at this question for quite a while now and it's obvious that I'm missing something blatant but can't find what!

Q. One stone is thrown vertically upwards with a speed of 2m/s and another is thrown vertically downwards with a speed of 2m/s. Both are thrown at the same time from a window 5m above the ground.

a/ which stone hits the ground first?
This one, the answer is obvious, the one thrown downwards because it travels a shorter distance with the same acceleration and starting velocity, but I can't prove it!

b/ Which stone is travelling fastest when it hits the ground?
I've done this one using v^2=u^2+2as, both equal 10.1m/s.

c/ What is the total distance travelled by each stone?
I'm guessing I need my answer (or lack of) from part a to work this out.

2. well the total distance travelled by the stone thrown down is obviously 5. try working out the total distance upwards the stone thrown upward travels (when U=2, V=0, a=-9.8 & s=?) and then that + the distance from this point to the ground. so (the distance upwards it travels x 2)+5
3. (Original post by indifferencepersonified)
Evening maths people, I've been looking at this question for quite a while now and it's obvious that I'm missing something blatant but can't find what!

Q. One stone is thrown vertically upwards with a speed of 2m/s and another is thrown vertically downwards with a speed of 2m/s. Both are thrown at the same time from a window 5m above the ground.

a/ which stone hits the ground first?
This one, the answer is obvious, the one thrown downwards because it travels a shorter distance with the same acceleration and starting velocity, but I can't prove it!

b/ Which stone is travelling fastest when it hits the ground?
I've done this one using v^2=u^2+2as, both equal 10.1m/s.

c/ What is the total distance travelled by each stone?
I'm guessing I need my answer (or lack of) from part a to work this out.

Work out distance using Suvat equations. For the first stone you have the following information. (I'm denoting downwards as positive)that S = 5m U = -2 V = ? A = 9.8 ms/2 T =?

For a use S = Ut + 1/2 (at^2) using the info I've just given you then for the second one it's the same except U is positive 2. Remember S = DISPLACEMENT.

For C Just work out the displacement of the first stone untill it's highest point, which is when final speed is 0. Double this then add the extra 5 m it falls, then another 5m stone two falls.
4. (Original post by robinyourpersie)
Work out distance using Suvat equations. For the first stone you have the following information. (I'm denoting downwards as positive)that S = 5m U = -2 V = ? A = 9.8 ms/2 T =?

For a use S = Ut + 1/2 (at^2) using the info I've just given you then for the second one it's the same except U is positive 2. Remember S = DISPLACEMENT.

For C Just work out the displacement of the first stone untill it's highest point, which is when final speed is 0. Double this then add the extra 5 m it falls, then another 5m stone two falls.
When I do part A for the time (of the stone thrown upwards) I get a quadraitc 4.9t^2+2t-5= 0 which will presumably give me two answers if I use the quadratic formula. I've gone through this before and it just never seemed right, and it never was right.
5. (Original post by indifferencepersonified)
Evening maths people, I've been looking at this question for quite a while now and it's obvious that I'm missing something blatant but can't find what!

Q. One stone is thrown vertically upwards with a speed of 2m/s and another is thrown vertically downwards with a speed of 2m/s. Both are thrown at the same time from a window 5m above the ground.

a/ which stone hits the ground first?
This one, the answer is obvious, the one thrown downwards because it travels a shorter distance with the same acceleration and starting velocity, but I can't prove it!

b/ Which stone is travelling fastest when it hits the ground?
I've done this one using v^2=u^2+2as, both equal 10.1m/s.

c/ What is the total distance travelled by each stone?
I'm guessing I need my answer (or lack of) from part a to work this out.

a/ Think about it!!! The one thrown upwards starts off at initial speed U (or is it V? Been a while since I done this stuff) and experiences acceleration of -9.81m/s/s due to gravity. For a brief moment it has velocity = 0 then it accelerates downwards, whereas the one thrown downwards has only to travel 5m downwards.

b/ Are you sure that's right? Double check it, remembering to take into account the differences in displacements.

c/ Total distance travelled by the one thrown downwards will be 5m. The one thrown upwards will be the distance it takes to reach its peak in the air, plus that distance again with 5m added.

EDIT: if you're getting quadratics at this level of physics then I think you are doing something wrong.
6. (Original post by Smack)
a/ Think about it!!! The one thrown upwards starts off at initial speed U (or is it V? Been a while since I done this stuff) and experiences acceleration of -9.81m/s/s due to gravity. For a brief moment it has velocity = 0 then it accelerates downwards, whereas the one thrown downwards has only to travel 5m downwards.

b/ Are you sure that's right? Double check it, remembering to take into account the differences in displacements.

c/ Total distance travelled by the one thrown downwards will be 5m. The one thrown upwards will be the distance it takes to reach its peak in the air, plus that distance again with 5m added.

EDIT: if you're getting quadratics at this level of physics then I think you are doing something wrong.
Believe it or not, I am thinking about it, I don't have a teacher so it's hard for me to spot errors when they come about. Yes I'm sure about b, the answer is the same as in the back of the book. I don't know how to work out how far upwards the stone is thrown upwards in order to find the distance it travels.
7. Acceleration = - 9.81
Starting speed = 5
Finishing speed = 0
Displacement = ???
8. surely u=2? and using s= (v^2-u^2)/2a I get 0.204m whi9ch presumably I would times by 2 because it comes back down again, then add the 5. That gives the right answer. I don't know why I kept going wrong. Thanks people. Oh and smack, in your first post you said check b, but the overall displacement of both balls is the same.
9. Yes, it is 2, that was my mistake - I was getting some of the given values muddled up.
10. for the downward ball, you should get quadratic, use the quadratic formula to solve it, you should get two values, one positive, the other negative, ignore the negative value as time cannot be negative in this case

for the other ball you divide calculating the time in to two stages, the upward motion and downward motion for the upward motion of the ball:

s=?
u=2m/s
v=0
a=-9.8
t=?
from this we can calculate the time by using the equation v=u+at, and we can also calculate the distance using the formula (v^2)=(u^2)+2as (which will help us later on). as now you have t and s

now for the downward motion
now to calculate t for the downward motion, you have
S=...
U=0
v=(not important)
a=9.8
t=?
YOu can get the value of s by adding the distance calculated for the upward motion and 5
then you use the formula s=ut+1/2at^2. and you will get quadratic. Then use the quadratic formula, as before their should be one positive root and one negative root. add the time for for downward motion and upward motion and you should get the total time

now if you compare the time for the two balls it should be obvious one is larger then the other, hence the ball with the shorter time is the one which hits the ground first
11. If you haven't already got this sorted I've just done some solutions for you, just about to scan them.

Edit: here they are, hope they help.
Attached Images

12. Yeh you will get two solutions, a negative one you will ignore and the solution you are after

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