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M1 Question - Suvat

A stone is thrown vertically upward from a point which is 3m above the ground, with speed 16ms^-1. Find a) the time of flight of the stone, b) the total distance travelled by the stone.

I know the answer to A and B.. My question is about B though.

B: s=? v=0 u=16 a=-9.8

v2=u2+2as[br]0=162+19.6s[br]s=13.06[br](13.06)2+3=29.12[br] v^2=u^2 + 2as[br]0=16^2 + 19.6s[br]s = 13.06[br](13.06)2 + 3 = 29.12[br]
Why is v = 0
And also why do you times by 2 and add 3.. I'm guessing what i worked out is for the first part of the journey when it goes to its maximum point, then i times it by 2 to get the other half. Then it falls 3m to the ground so i add 3? Oh so v = 0 because it is when it reaches the maximum height it stops to turn round?
I just realised why as i typed it, but ill post it anyway just to make sure i have the correct understanding
(edited 7 years ago)
Original post by AdeptDz
A stone is thrown vertically upward from a point which is 3m above the ground, with speed 16ms^-1. Find a) the time of flight of the stone, b) the total distance travelled by the stone.I know the answer to A and B.. My question is about B though.B: s=? v=0 u=16 a=-9.8 v^2=u^2 + 2as0=16^2 + 19.6ss = 13.06(13.06)2 + 3 = 29.12[\tex]Why is v = 0And also why do you times by 2 and add 3.. I'm guessing what i worked out is for the first part of the journey when it goes to its maximum point, then i times it by 2 to get the other half. Then it falls 3m to the ground so i add 3? Oh so v = 0 because it is when it reaches the maximum height it stops to turn round? I just realised why as i typed it, but ill post it anyway just to make sure i have the correct understanding
v = 0 because the stone decelerates and instantaneously comes to rest (at its maximum point) and then falls back downyou double it and add 3 because when it falls back down it travels the original distance plus an extra 3 meters since it was thrown initially 3 meters above the ground.
I only learnt this today but I can explain. v = 0 because the ball hits the ground and stops moving. The end velocity could also be -16m/s as it is going the same speed but in the other direction (down) but because you chose to use 0 instead you need to times your answer by 2 to make it equivalent to what it would be if you used the minus starting velocity.

I hope that that helped. Ask if it didn't please.
Original post by AdeptDz
...


v=0 at maximum height because vertically the stone is not moving up nor down at that instant.

For other one, because it goes up (one distance), then goes down to the spot where it came from (the same distance again), and falls an extra 3m to hit the ground.
(edited 7 years ago)
Reply 4
Original post by AspiringUnderdog
I only learnt this today but I can explain. v = 0 because the ball hits the ground and stops moving. The end velocity could also be -16m/s as it is going the same speed but in the other direction (down) but because you chose to use 0 instead you need to times your answer by 2 to make it equivalent to what it would be if you used the minus starting velocity.

I hope that that helped. Ask if it didn't please.


Original post by BasharAssad
v = 0 because the stone decelerates and instantaneously comes to rest (at its maximum point) and then falls back downyou double it and add 3 because when it falls back down it travels the original distance plus an extra 3 meters since it was thrown initially 3 meters above the ground.


Thanks, i got it :grin:
Original post by AspiringUnderdog
I only learnt this today but I can explain. v = 0 because the ball hits the ground and stops moving. The end velocity could also be -16m/s as it is going the same speed but in the other direction (down) but because you chose to use 0 instead you need to times your answer by 2 to make it equivalent to what it would be if you used the minus starting velocity.

I hope that that helped. Ask if it didn't please.


No. v=0 for reason above. The velocity at the instant it hits the ground is definitely not -16m/s either, it is -16m/s at the instant it reaches its starting point.

While you are correct about v=0 at the instant stone hits the ground and begins to rebound, that information is not as useful.
(edited 7 years ago)
Original post by RDKGames
No. v=0 for reason above. The velocity at the instant it hits the ground is definitely not -16m/s either, it is -16m/s at the instant it reaches its starting point.


It is-16m/s right up to the end point which is a tiny a fraction of a second difference so it can be considered to be the final speed.
Original post by AspiringUnderdog
It is-16m/s right up to the end point which is a tiny a fraction of a second difference so it can be considered to be the final speed.


No it's not. The 'end point' would be the ground (unless stated otherwise by the question), and the speed at which the stone hits the ground is 17.7 m/s to 3s.f. which is not anywhere near 16.

Taking up as the positive direction, and going down from the starting point, we have:
v2=u2+2as=(16)2+2(9.8)(3)=314.8v^2=u^2+2as=(-16)^2+2(-9.8)(-3)=314.8 therefore v=17.7426....v=17.7426....
(edited 7 years ago)
Reply 8
Original post by AspiringUnderdog
I only learnt this today but I can explain. v = 0 because the ball hits the ground and stops moving. The end velocity could also be -16m/s as it is going the same speed but in the other direction (down) but because you chose to use 0 instead you need to times your answer by 2 to make it equivalent to what it would be if you used the minus starting velocity.

I hope that that helped. Ask if it didn't please.


Which is it? Is the final velocity 0m/s or -16m/s?

RDKGAmes's reply immediately above resolves the issue.
Reply 9
Original post by AspiringUnderdog
I only learnt this today but I can explain.

I hope that that helped. Ask if it didn't please.


It didn't - I get that you only learnt this today and want to explain, but if you're unsure about things it's better to not post anything than post the wrong thing as you have here - especially with a student who is learning at a beginner level who won't know you're wrong. Hope you understand, thanks!
Yh, you have got it, and when v = 0 is because at the top before it falls it has no velocity at its max point.
Reply 11
Thanks for all the input ^


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