A ball is thrown vertically upwards at 19.6 m/s. Taking the acceleration due to gravity as 9.8 m/s per second and ignoring air resistance, how high above the throwing point does it rise?

1. Choose upwards as the positive direction, so the acceleration is minus 9.8 metres per second per second. 2. The initial velocity is plus 19.6 metres per second and the velocity at the highest point is zero. 3. Use the equation of motion that links final velocity, initial velocity, acceleration and displacement. 4. Zero squared equals 19.6 squared plus two multiplied by minus 9.8 multiplied by the displacement. 5. Nineteen point six squared is 384.16 metres squared per second squared, and two multiplied by 9.8 is 19.6 metres per second per second. 6. So the displacement is 384.16 divided by 19.6, which is 19.6 metres upwards. 7. The 9.8 metre answer traps anyone who divides the initial velocity by the acceleration and calls the result a height. 8. The 39.2 metre answer doubles the correct height by leaving out the factor of two in the equation, and the 4.9 metre answer uses the value of half of the acceleration as a distance. _Source: Siyavula Physical Sciences Grade 12 (Everything Science, CC BY 4.0), Chapter 3: Vertical projectile motion in one dimension_