m1 help :(

kiiten

Different questions. Help please

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Reply 1

TeeEm

Original post by kiiten

Different questions. Help please

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any workings?

Reply 2

kiiten

For ques 3. I guessed s=3 and a=-9.8 i put that into s=ut+1/2at^2 and got 7.9=t (u+1/2t)

For ques 1. I wasnt sure because i had the sane problem as before (too many unknowns)

Reply 3

samb1234

Original post by kiiten

For ques 3. I guessed s=3 and a=-9.8 i put that into s=ut+1/2at^2 and got 7.9=t (u+1/2t)

For ques 1. I wasnt sure because i had the sane problem as before (too many unknowns)

What can you say about the value of the J vector when it hits the ground, and how can we use this to find t

Reply 4

kiiten

Original post by samb1234

What can you say about the value of the J vector when it hits the ground, and how can we use this to find t

Ah j becomes 0 so do you solve 6+9.5t-5t^2=0 to find t? If so could you x by -1 to make the 5 +ve?

Reply 5

samb1234

Original post by kiiten

Ah j becomes 0 so do you solve 6+9.5t-5t^2=0 to find t? If so could you x by -1 to make the 5 +ve?

Yes

What about ques. 3??

Wait, how would you plot the path if you only have one value for t

Anyone? :,(

Original post by kiiten

Anyone? :,(

did you want help with q 1 or 3?
i haven't done i and j notation yet but i have done suvat

Reply 10

thefatone

Original post by kiiten

For ques 3. I guessed s=3 and a=-9.8 i put that into s=ut+1/2at^2 and got 7.9=t (u+1/2t)

For ques 1. I wasnt sure because i had the sane problem as before (too many unknowns)

lol wat where did u get time from?
I'll define up as positive

If the max height the stone reaches is 5 m and the stone is thrown from 2 m above the ground the the distance it travels is 3m until it reaches its max height
so s=3 which you got
If up is positive and acceleration due to gravity acts down then
a= -9.8
u=u
since the stone will stop momentarily for a split second at its max height
v=0
i'm sure you can work out the rest from there

part b
defining positive as down
so on your diagram you'll be looking at the right hand side which has 5m labelled on the long line
so finding the velocity of the stone when it hits the ground
Remember now u=0 because we start by calculating from when the stone is momentarily still for a split second
v=v
a=9.8
s=5
i'm sure you can finish that one too

part c
try and do this one by yourself
split the diagram into 2 where you split the diagram vertically at the point where the stone is momentarily still
calculate the time for the first bit before the stone reaches max height and calculate time for after the stone has reached its max height and has reached the ground
add them together and you should have your answer

(edited 8 years ago)

Reply 11

kiiten

Original post by thefatone

did you want help with q 1 or 3?
i haven't done i and j notation yet but i have done suvat

Ok, ques 3. please - still need help on ques 1 though (for anyone else)

Original post by kiiten

Ok, ques 3. please - still need help on ques 1 though (for anyone else)

Question 3 is just suvat, question 1 you just need to use the value of t you found in part a

Reply 14

kiiten

Original post by samb1234

Question 3 is just suvat, question 1 you just need to use the value of t you found in part a

Im confused for part b on ques 3. Does s become 6? Also, for ques 1 you will only get 1 value so how do you plot a graph or are you supposed to do it when t is 1,2,3, etc.

Reply 15

samb1234

No s would be 5. As far as plotting the path its just going to be a parabola, just plug some numbers in if you want a more accurate picture.

Original post by kiiten

Im confused for part b on ques 3. Does s become 6? Also, for ques 1 you will only get 1 value so how do you plot a graph or are you supposed to do it when t is 1,2,3, etc.

Reply 16

kiiten

Original post by samb1234

No s would be 5. As far as plotting the path its just going to be a parabola, just plug some numbers in if you want a more accurate picture.

For the suvat im confused on part b. Could someone please explain. Do you use v^2 = u^2 + 2as and if you do s=5 and u=0 (5 is max. height) you get a -ve

Posted from TSR Mobile

Reply 17

Zacken

Original post by thefatone

@Zacken

Hey.

Original post by kiiten

For the suvat im confused on part b. Could someone please explain. Do you use v^2 = u^2 + 2as and if you do s=5 and u=0 (5 is max. height) you get a -ve

Posted from TSR Mobile

Plot the curve, just draw a parabola. There's no need for suvat in part b.

Unless you're talking about "velocity of a stone as it hits the ground, in which case s=-2."

(edited 8 years ago)

Reply 18

kiiten

Original post by Zacken

Hey.

Plot the curve, just draw a parabola. There's no need for suvat in part b.

Unless you're talking about "velocity of a stone as it hits the ground, in which case s=-2."

So just use values of t like 1, 2 etc.

Yes, it says velocity when the stone hits the ground. If you use suvat why would s=-2 dont you have to account for the distance from when its thrown up?

Reply 19

Zacken

Original post by kiiten

Yes, it says velocity when the stone hits the ground. If you use suvat why would s=-2 dont you have to account for the distance from when its thrown up?