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
533 exam-style questions on CIE IGCSE Physics 1.7 Energy, work and power, covering 1.7.1a Energy stores and energy transfers, 1.7.1b Conservation of energy, 1.7.1c Kinetic and gravitational potential energy, 1.7.2 Work done, 1.7.3 Energy resources, 1.7.3a Energy resources and electricity generation, 1.7.3b Nuclear fusion as an energy source, 1.7.3c Efficiency of energy transfer, and 1.7.4 Power. Each one has a worked solution and a mark scheme showing where the marks go.