Higher November 2017 Paper 2 Q20

20 A stone is thrown upwards with a speed of \(v\) metres per second.

The stone reaches a maximum height of \(h\) metres.

\(h\) is directly proportional to \(v^2\)

When \(v = 10\), \(h = 5\)

Work out the maximum height reached when \(v = 24\) [4 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
\(h = kv^2\) or \(5 = k \times 10^2\) or \(5 \div 10^2\) or \(5 : 10^2\)	M1	oe
(\(k =\)) \(\dfrac{1}{20}\) or (\(k =\)) 0.05 or \(h = \dfrac{1}{20}v^2\) or \(h = 0.05v^2\)	A1	oe Correct value for \(k\) or correct equation in \(h\) and \(v\)
their \(\dfrac{1}{20} \times 24^2\)	M1dep	oe \(\dfrac{1}{20} \times 24^2\) implies M1A1M1
28.8	A1ft	ft their \(k\) and M1A0M1
Alternative method 2
\(kh = v^2\) or \(k \times 5 = 10^2\) or \(10^2 \div 5\) or \(10^2 : 5\)	M1	oe
(\(k =\)) 20 or \(20h = v^2\)	A1	oe Correct value for \(k\) or correct equation or correct equation in \(h\) and \(v\)
\(24^2 \div\) their 20	M1dep	oe \(24^2 \div 20\) implies M1A1M1
28.8	A1ft	ft their \(k\) and M1A0M1
Alternative method 3
\(\left(\dfrac{24}{10}\right)^2\) or \(\dfrac{576}{100}\) or \(24^2 : 10^2\)	M1	oe
\(\dfrac{h}{5} = \left(\dfrac{24}{10}\right)^2\)	A1	oe Correct equation in \(h\)
\(5 \times\) their \(\left(\dfrac{24}{10}\right)^2\)	M1dep	oe \(5 \times\) \(\left(\dfrac{24}{10}\right)^2\) implies M1A1M1
28.8	A1ft	ft their \(\left(\dfrac{24}{10}\right)^2\) and M1A0M1
Alternative method 4
\(\left(\dfrac{10}{24}\right)^2\) or \(\dfrac{100}{576}\) or \(10^2 : 24^2\)	M1	oe
\(\dfrac{5}{h} = \left(\dfrac{10}{24}\right)^2\)	A1	oe Correct equation in \(h\)
\(5 \div\) their \(\left(\dfrac{10}{24}\right)^2\)	M1dep	oe \(5 \div\) \(\left(\dfrac{10}{24}\right)^2\) implies M1A1M1
28.8	A1ft	ft their \(\left(\dfrac{24}{10}\right)^2\) and M1A0M1

\(h \propto v^2\) with no further valid working	Zero
\(h = kv\) or \(h = kv^3\) or \(h = \dfrac{k}{v^2}\) etc not recovered	Zero
Up to first two marks can be awarded for correct working even if not subsequently used
Allow use of other letters