M2 June 2006 Q5b
Maths and statistics discussion, revision, exam and homework help.
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Re: M2 June 2006 Q5bVertically for A, the displacement of the stone at time t, relative to the top of the cliff, is -gt^2, since u is zero.(Original post by lekha2611)
For part b, why in the markscheme (attached), does it say that the distance moved by A is 73.5-1/2gt^2? I appreciate that the particle A is falling from a height, but why did the 73.5 appear in the SUVAT equation s=ut+1/2at^2, in the place of ut?
Thanks
So the height relative to the bottom of the cliff is 73.5-gt^2
And you need both heights relative to the same datum (the bottom of the cliff in this case), as you're going to be equating them and solving for t. -
Re: M2 June 2006 Q5bOh okay, would that be hinted in the question, or is it something I'll just need to remember for similar questions?(Original post by ghostwalker)
Vertically for A, the displacement of the stone at time t, relative to the top of the cliff, is -gt^2, since u is zero.
So the height relative to the bottom of the cliff is 73.5-gt^2
And you need both heights relative to the same datum (the bottom of the cliff in this case), as you're going to be equating them and solving for t. -
Re: M2 June 2006 Q5bI doubt if it would be hinted at.(Original post by lekha2611)
Oh okay, would that be hinted in the question, or is it something I'll just need to remember for similar questions?
The fact that the two particle/masses are leaving from two different places is the clue that you're going to have to do something in order to be able to compare their vertical positions. -
Re: M2 June 2006 Q5bah right I'll remember that in the future then - thank you for your help!(Original post by ghostwalker)
I doubt if it would be hinted at.
The fact that the two particle/masses are leaving from two different places is the clue that you're going to have to do something in order to be able to compare their vertical positions.