(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

http://www.physics.ox.ac.uk/olympiad/Downloads/PastPapers/BPhO_PC_2006_QP.pdf

Question 11

2. Relevant equations

h=v^2/g

3. The attempt at a solution

convert 36km/h into 10m/s

10 is final velocity so average should be 5m/s

h=v^2/g=5^2/10=2.5m

Using pythagoras theorem we can calculate that the other length must be roughly 50

(2.5^2+x^2=50^2)

Therefore since the gradient is the change in y over the change in x,

shouldn't the gradient just be 2.5/50=0.05? The actual answer is 0.101, they say the word 'tangent' in the markscheme but offer no explanation for it and I do not understand it.

Any help would be great, thank you very much

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# Calculating gradient with distance and speed

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