jamestg Maths help

Why is sin being used?! This hasn't come up in any of our lessons!

I've just taken a look at your working and it's all wrong.

v\neq 0

when it lands on the ground, so your equation is invalid, s=+ve also invalid.

Just consider the total displacement from the beginning to the end, it's s=0, u=30, a=-9.8, find t from this.

Reply 21

jamestg

OP

21

Original post by Zacken

I've just taken a look at your working and it's all wrong.

v\neq 0

when it lands on the ground, so your equation is invalid, s=+ve also invalid.

Just consider the total displacement from the beginning to the end, it's s=0, u=30, a=-9.8, find t from this.

Thank you for being straightforward! I'd much rather that than being nice!

So the displacement is 0 because it returns to its original position? I got 6.1secs...

From:

t(4.9t-30)=0
4.9t-30=0
and then that being rearranged

Reply 22

Zacken

22

Original post by jamestg

Thank you for being straightforward! I'd much rather that than being nice!

So the displacement is 0 because it returns to its original position? I got 6.1secs...

From:

t(4.9t-30)=0
4.9t-30=0
and then that being rearranged

Haha, yeah, I'm being a little abrupt, sorry - it's 2 in the morning and I'm trying to hurry to bed at the same time. :lol:

That's correct, 0 because it returns to it's original position and your answer of t=6 something is correct. :-)

Reply 23

jamestg

OP

21

Original post by Zacken

Haha, yeah, I'm being a little abrupt, sorry - it's 2 in the morning and I'm trying to hurry to bed at the same time. :lol:

That's correct, 0 because it returns to it's original position and your answer of t=6 something is correct. :-)

Hahahah it's okay, being harsh is a better motivator than being nice tbh.

Thank you for your help and hopefully you can get to sleep now, sorry if I bothered you!

Reply 24

Zacken

22

Original post by jamestg

Hahahah it's okay, being harsh is a better motivator than being nice tbh.

Thank you for your help and hopefully you can get to sleep now, sorry if I bothered you!

Great! Glad I was of some help. :-)

You weren't, 'course not, voluntary help, after all. :wink:

Night!

Reply 25

jamestg

OP

21

A stone is thrown vertically upwards with speed 16 m s–1 from a point h metres above theground. The stone hits the ground 4 s later. Find (a) the value of h,
(b) the speed of the stone as it hits the ground. For part (a) I got

s=h u=16 v=0 a=-9.8 t=4

therefore I'll use the equation s=ut+0.5at²

h=(16x4)+(0.5*-9.8*16)
h=64+-78.4
h=-14.4

h is therefore 14.4m

For part (b) I got

V=u+at

V=? u=16 a=-9.8 t=4

V=16+(-9.8*4)
V= 23.2ms^-1

But why is acceleration negative when it's going down? Is it because it's overall, as the time is accounting for when it is both going up and down and the initial velocity used is right at the beginning of the stone's path?

Reply 26

16Characters....

13

Original post by jamestg

But why is acceleration negative when it's going down? Is it because it's overall, as the time is accounting for when it is both going up and down and the initial velocity used is right at the beginning of the stone's path?

Any object moving in Earth's gravitational field close to the surface without resistance accelerates vertically downwards at 9.8 ms^-2. In your example since you are treating the vertically upwards direction as positive this means you are using a = -9.8ms^-2 vertically upwards. This acceleration is constant. The gravitational acceleration is independent of the direction in which object is moving.

Reply 27

jamestg

OP

21

What does "express these as square roots of integers" actually mean.

For example - 3(root)5

Reply 28

Pronged Lily

8

Original post by jamestg

What does "express these as square roots of integers" actually mean.

For example - 3(root)5

Write that as a single square root.
Like 2sqr3 = sqr12

Reply 29

jamestg

OP

21

Original post by Pronged Lily

Write that as a single square root.
Like 2sqr3 = sqr12

Cracking, thanks!

Reply 30

pereira325

20

lol just saw this. it's great you have your own thread though :smile:

brilliant idea, might copy if I can be bothered!

Reply 31

Louisb19

14

Original post by Zacken

...

Do you know if my answer in that integration thread from earlier correct? It had

\ln(1+k)

in it.

Reply 32

Zacken

22

Original post by Louisb19

Do you know if my answer in that integration thread from earlier correct? It had

\ln(1+k)

in it.

If your answer was

\dfrac{\pi^2}{12}

then it's fine.

Reply 33

jamestg

OP

21

Original post by pereira325

lol just saw this. it's great you have your own thread though :smile:

brilliant idea, might copy if I can be bothered!

I would do it - stops the maths section from being clogged up!

Also, when it comes to revision you can look back through all the stuff you didn't understand during the year and you can check whether or not you now understand it!

Reply 34

jamestg

OP

21

Please can someone guide me in the right direction with this dynamics question?

I've hit a brick wall lol.

That's the question

Attachment not found

That's what I've got so far

Thank you!!

Original post by jamestg

Please can someone guide me in the right direction with this dynamics question?

I've hit a brick wall lol.

That's the question

Attachment not found

That's what I've got so far

Thank you!!

You seem to be assuming that there is some force T that is continually pushing the particle up as it rises through the water.

If you throw a ball into the air, there was an initial force you applied to the ball but only gravity is acting on it as it rises through the air.

It's the same here : there are only downwards forces acting on the particle as it goes up.

Without T, this should make your life easier.

(edited 8 years ago)

Reply 36

math42

20

Original post by jamestg

Please can someone guide me in the right direction with this dynamics question?

I've hit a brick wall lol.

That's the question

Attachment not found

That's what I've got so far

Thank you!!

If you saw my post before ignore it I can't read and/or think

Reply 37

jamestg

OP

21

Original post by 1 8 13 20 42

If you saw my post before ignore it I can't read and/or think

Hahaha okay! I feel you :redface:

Reply 38

jamestg

OP

21

Original post by notnek

You seem to be assuming that there is some force T that is continually pushing the particle up as it rises through the water.

If you throw a ball into the air, there was an initial force you applied to the ball but only gravity is acting on it as it rises through the air.

It's the same here : there are only downwards forces acting on the particle as it goes up.

Without T, this should make your life easier.

I get what you're saying overall with the continual pushing force, but I'm not entirely sure now how to go about the question...

Surely I must use T still? What other equation could I do? :frown:

Reply 39

math42

20

Original post by jamestg

Hahaha okay! I feel you :redface:

lol I think I know what's going on now, won't make any promises though
Can you not use v^2 = u^2 + 2as for finding the height? Note that as the other person said there are only downwards forces; these can be used to find resultant acceleration

jamestg Maths help

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