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    Can someone please help with the first question, worth 6 marks
    Attachment 728380

    What I did:
    Found the resultant velocity \sqrt{30^2+100^2}=104.4
    Energy at A: \frac{1}{2}m(104.4)^2

    Energy at B: \frac{1}{2}mv^2+mg(20)

    After equating energy at A with energy at B, I was able to cancel out the mass, I rearranged to get V at B as 102.5m/s
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    (Original post by joyoustele)
    Can someone please help with the first question, worth 6 marks
    Attachment 728380
    Have you made any progress?
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    (Original post by Notnek)
    Have you made any progress?
    Yes, But I think its compelety wrong, from the start.
    Im posting it now

    Could I have a tip?
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    (Original post by joyoustele)
    Can someone please help with the first question, worth 6 marks
    Attachment 728380

    What I did:
    Found the resultant velocity \sqrt{30^2+100^2}=104.4
    Energy at A: \frac{1}{2}m(104.4)^2

    Energy at B: \frac{1}{2}mv^2+mg(20)

    Is this correct to start off with?
    This looks good, next set them equal to each other because of conservation of energy.

    By the way there's no need to find the square root of 30^2 + 100^2 because you'll have to square it anyway in the equation.
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    (Original post by joyoustele)
    Can someone please help with the first question, worth 6 marks
    Attachment 728380

    What I did:
    Found the resultant velocity \sqrt{30^2+100^2}=104.4
    Energy at A: \frac{1}{2}m(104.4)^2

    Energy at B: \frac{1}{2}mv^2+mg(20)

    After equating energy at A with energy at B, I was able to cancel out the mass, I rearranged to get V at B as 102.5m/s
    This is correct so now you have the speed at B.

    Next notice that the horizontal component of the velocity will remain the same (since there's no horizontal acceleration) so it will be 100 at B.

    So you have the horizontal component of the velocity at B and you have the speed - can you use this to find the vertical component and hence find the velocity at B?
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    (Original post by Notnek)
    This is correct so now you have the speed at B.

    Next notice that the horizontal component of the velocity will remain the same (since there's no horizontal acceleration) so it will be 100 at B.

    So you have the horizontal component of the velocity at B and you have the speed - can you use this to find the vertical component and hence find the velocity at B?
    Thanks a lot notnek
 
 
 
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