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M1 Projectiles

If I don't get part A correct, I'm screwed for the rest of the questions.

Is it tan inverse 6/8 ?? ie 36.9 degrees? (3SF)

IMG-20120506-00122.jpg
Reply 1
Where do you get the 8 from? Use toa.
Reply 2
Original post by Hopple
Where do you get the 8 from? Use toa.


So its tan inverse 6/10?

IDK but I put the velocity of 8 as part of the triangle :s-smilie: then assumed 10 must have been the hyptonuse, so i thought 8 was the adjacent :|
Reply 3
Original post by MSI_10
So its tan inverse 6/10?
Yeah, that's it.
Reply 4
yes tanα is 6/10
Reply 5
Original post by Hopple
Yeah, that's it.


Aight thanks.

Am I starting part b probably; to get time, first need to use S. Do I use 6 or 10..?

Then SUVAT (S=ut+o.5at^2) use quadratic formula to get t?
Reply 6
Original post by MSI_10
Aight thanks.

Am I starting part b probably; to get time, first need to use S. Do I use 6 or 10..?

Then SUVAT (S=ut+o.5at^2) use quadratic formula to get t?


For part b, are you looking at vertical or horizontal motion (or both)? And yes, that suvat equation will do.
Reply 7
Original post by Hopple
For part b, are you looking at vertical or horizontal motion (or both)? And yes, that suvat equation will do.


I did this..
S=-6
a=-9.8
U=8sin31..

-6=8sin31t-4.9t^2
-4.9^2+8sin31t+6=0
got t=1.60 or t=-0.763

Answer t=1.60..?
Reply 8
Original post by MSI_10
I did this..
S=-6
a=-9.8
U=8sin31..

-6=8sin31t-4.9t^2
-4.9^2+8sin31t+6=0
got t=1.60 or t=-0.763

Answer t=1.60..?


Make sure you're measuring everything in the same direction.
Reply 9
Original post by Hopple
Make sure you're measuring everything in the same direction.


:s-smilie: :\
Reply 10
Original post by MSI_10
:s-smilie: :\


Even though everything's going downwards, you've made displacement and acceleration be in the opposite direction to initial velocity.
Reply 11
Original post by Hopple
Even though everything's going downwards, you've made displacement and acceleration be in the opposite direction to initial velocity.


Oh that's what you meant, yeah i get what you mean now thanks:smile:

So t=0.763
Reply 12
Hmm, can anyone check if im doing part c correctly.

It's asking for distance.. do I use the time obtained from part b, use S=ut+1/2at^2 but use 8cos31 for u instead of 8sin31..?
Reply 13
Original post by MSI_10
Hmm, can anyone check if im doing part c correctly.

It's asking for distance.. do I use the time obtained from part b, use S=ut+1/2at^2 but use 8cos31 for u instead of 8sin31..?


Yes, but remember there are no forces acting horizontally here.
Reply 14
Original post by finality
Yes, but remember there are no forces acting horizontally here.



Alright so I used 8sin31
I get the answer as 6m. Something tell's me this aint right; it seems to high :\

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