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    (Original post by Me123456789)
    So would I use

    s = 1/2(u+v)t

    where t=2.7, s=35.7 (my answer to part b) and u=0 ?
    I'm not really sure where you got t=2.7 or s=35.7 from?

    solving vertically: u=0 a=3 (from part b) and t=0.9 so now you can find v for vertical component of v

    solving horizontally: u=0.52 a=0 and t=0.9 so you can find v for horizontal motion.

    Then use Pythagros to find resultant of two velocities to find speed
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    (Original post by Username_valid)
    I'm not really sure where you got t=2.7 or s=35.7 from?

    solving vertically: u=0 a=3 (from part b) and t=0.9 so now you can find v for vertical component of v

    solving horizontally: u=0.52 a=0 and t=0.9 so you can find v for horizontal motion.

    Then use Pythagros to find resultant of two velocities to find speed
    Oh no sorry, I'm talking about part c of the question in the attached image?
    Name:  2013-03-20 22.17.08.jpg
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    (Original post by Me123456789)
    Oh no sorry, I'm talking about part c of the question in the attached image?
    Name:  2013-03-20 22.17.08.jpg
Views: 36
Size:  266.8 KB
    Oh right sorry, in that case yes I think you're right. But then you need to find the resultant of that answer and you're answer from part a to find the speed
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    (Original post by Username_valid)
    Oh right sorry, in that case yes I think you're right. But then you need to find the resultant of that answer and you're answer from part a to find the speed
    For part a my answer is 21.5 ms and for part b it is 35.7 m, knowing that t=2.7 what would be the simplest equation for me to use?
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    (Original post by Me123456789)
    For part a my answer is 21.5 ms and for part b it is 35.7 m, knowing that t=2.7 what would be the simplest equation for me to use?
    What you said previously was correct:

    (Original post by Me123456789)
    So would I use

    s = 1/2(u+v)t

    where t=2.7, s=35.7 (my answer to part b) and u=0 ?
    Solving vertically a=9.8 m.s^-2 so you could of also used v=u+at to give you v=0+ 9.8*2.7 and this would give you the same answer as above.

    This gives you the vertical component of v. For the horizontal component you could simply use your answer from part a (21.48m.s^-1) because a=0 so applying the formula v=u+at you yould get V=21.48 + (0*2.7).

    So now you have the final velocity for the vertical and horizontal components which is sides a and b on the image Name:  pythagoras-theorem.gif
Views: 26
Size:  1.8 KB

    To find speed, you have to find the resultant of the two components, so find side c. Quote me for any more help and is this for M1?
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    (Original post by Username_valid)
    What you said previously was correct:



    Solving vertically a=9.8 m.s^-2 so you could of also used v=u+at to give you v=0+ 9.8*2.7 and this would give you the same answer as above.

    This gives you the vertical component of v. For the horizontal component you could simply use your answer from part a (21.48m.s^-1) because a=0 so applying the formula v=u+at you yould get V=21.48 + (0*2.7).

    So now you have the final velocity for the vertical and horizontal components which is sides a and b on the image Name:  pythagoras-theorem.gif
Views: 26
Size:  1.8 KB

    To find speed, you have to find the resultant of the two components, so find side c. Quote me for any more help and is this for M1?
    Thank you!

    No, it's unit 2 physics.
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    (Original post by Me123456789)
    Thank you!

    No, it's unit 2 physics.
    Ah OK, no problem. All the best
 
 
 
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