Question

An Olympic diver of mass 50 kg, initially at rest, dives vertically downward and it takes 2 s for her to touch the water. What would be her velocity and kinetic energy, respectively, just before she touches the water?

Answer 1

• Given:

  • m = 50 kg

  • t = 2 s

  • u = 0 m/s

  • a = 10 $m/s^2$

  • v =?

  • KE =?

Solve and explain:

• Finding the magnitude of final velocity of the diver:

v = u + at

= 0 + (10)*(2)

= 20 m/s

• Finding the kinetic energy of the diver just before touching the water:

$KE=½m{v}^2$
$KE=½\ast 50\ast {20}^2$

= 10 kJ