An arrow is shot vertically upwards at a speed of 6.0 m/s.
All of the energy initially in the kinetic store of the arrow is transferred to its gravitational potential store when the arrow reaches its maximum height.
Assume the gravitational field strength g g\,g is 9.8 m/s2.
Which equation gives the maximum height h h\,h reached by the arrow?
h=9.82×6.02h = \frac{9.8}{2 \times 6.0^2}h=2×6.029.8
h=6.02×29.8h = \frac{6.0^2 \times 2}{9.8}h=9.86.02×2
h=6.022×9.8h = \frac{6.0^2}{2 \times 9.8}h=2×9.86.02
h=2×9.86.02h = \frac{2 \times 9.8}{6.0^2}h=6.022×9.8
229 exam-style questions on CIE IGCSE Physics 1.7.1a Energy stores and energy transfers. Each one has a worked solution and a mark scheme showing where the marks go.