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# Motion in a vertical plan question Watch

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1. a stone is dropped from the top of a tower. 3 seconds later another stone is thrown downwards from the same point at 35m/s. if they hit the ground at the same time find the height of the tower.

This is technically a mechanics 1 question but im sure the physics forum will have no trouble
2. S = ?
u = 35 ms^-1
v =
a = 9.8 ms^-2 (because the particle is being dropped downwards we take gravity to be positive)
t = 3s

Now, use an equation that coincides with all of our measurements: in order to find s (the displacement).
3. (Original post by TarotOfMagic)
S = ?
u = 35 ms^-1
v =
a = 9.8 ms^-2 (because the particle is being dropped downwards we take gravity to be positive)
t = 3s

Now, use an equation that coincides with all of our measurements: in order to find s (the displacement).
I already tried that. Like this:

S = ut + 1/2at^2
S = 35 (3) + 4.9(3^2)
S = 105 + 44.1
S = 149.5 m

But according to the answer this is wrong it should be 579.5 m

4. (Original post by hals)
a stone is dropped from the top of a tower. 3 seconds later another stone is thrown downwards from the same point at 35m/s. if they hit the ground at the same time find the height of the tower.

This is technically a mechanics 1 question but im sure the physics forum will have no trouble
the struggle is real
5. (Original post by u.sha)
the struggle is real
Oh look we have matching avatars 😁
6. (Original post by TarotOfMagic)
S = ?
u = 35 ms^-1
v =
a = 9.8 ms^-2 (because the particle is being dropped downwards we take gravity to be positive)
t = 3s

Now, use an equation that coincides with all of our measurements: in order to find s (the displacement).
You're an idiot. The stone moving at 35 m/s is dropped 3 seconds after the first, it doesn't take 3 seconds to reach the floor.
(Original post by hals)
a stone is dropped from the top of a tower. 3 seconds later another stone is thrown downwards from the same point at 35 m/s. if they hit the ground at the same time find the height of the tower.

This is technically a mechanics 1 question but im sure the physics forum will have no trouble
For the first stone,
s=
u=0
v=
a=9.8
t=t+3

For the second stone
s=
u=35
v
a=9.8
t=t

Equate the two displacements to find time, then find displacement using that value of t
7. (Original post by an_atheist)
You're an idiot. The stone moving at 35 m/s is dropped 3 seconds after the first, it doesn't take 3 seconds to reach the floor.

For the first stone,
s=
u=0
v=
a=9.8
t=t+3

For the second stone
s=
u=35
v
a=9.8
t=t

Equate the two displacements to find time, then find displacement using that value of t
Thanks
Shouldn't t=t-3 for the 1st stone because it happens before the 2nd one -.-

P.s. That idiot remark was unnecessary but ill let it slide because you tried to help
8. (Original post by hals)
Thanks
Shouldn't t=t-3 for the 1st stone because it happens before the 2nd one -.-

P.s. That idiot remark was unnecessary but ill let it slide because you tried to help
**** you are correct. Sorry about the idiot thing, it annoys me when people (inclding myself) manage to mess up easy maths based problems
9. (Original post by an_atheist)
**** you are correct. Sorry about the idiot thing, it annoys me when people (inclding myself) manage to mess up easy maths based problems
Apology accepted lol. Obviously that 1st method i tried wouldnt have worked because t=3 was not the actual time of fall but then i didnt know wht else to do and came on here after trying that invalid method
10. (Original post by hals)
Apology accepted lol. Obviously that 1st method i tried wouldnt have worked because t=3 was not the actual time of fall but then i didnt know wht else to do and came on here after trying that invalid method
Nope, it is t+3 for the first stone. IF the second stone is falling for a time t, the first stone was released 3 seconds earlier, and was falling for t+3 seconds

Got 579.5m for the height of the tower.
11. (Original post by an_atheist)
**** you are correct. Sorry about the idiot thing, it annoys me when people (inclding myself) manage to mess up easy maths based problems
Irony, at its finest.

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