A 5 kg rock is dropped over the edge of a 15 m cliff. What is the momentum of the rock as it strikes the ground

Momentum of the rock = 85.7 Kgm/s

Explanation:

Given

Rock weight = 5 kg

height of a cliff = 15 m

find Momentum =?

  Momentum = mass × velocity

 we don't have velocity here so let's find the velocity v

Potential energy of the rock will be = mgh

  P.E. = mgh

 ⇒   5  × 9.8 × 15

 ⇒   735 J

 The Kinetic energy before a moment of touching the ground will be = maximum potential energy = 735 J   ⇒ Going by the law of conservation law.

 K.E. = 1/2 m × v²

 735 = 1/2 ×  5 ×  v²

    v² =   294

     v  = 17.14 m/s  

Now get on to the momentum

Momentum = m ×  v

Momentum = 5 ×  17.14

Momentum =  85.7  Kgm/s

The Momentum of the rock as it strikes ground will be 85.7 Kgm/s.

snell's law

refractve index = sin incidence / sin refraction