A3 kg stone is dropped from a height of 4 m what is its momentum as it strikes the ground

Okay first you have to recognize that the maximum Gravitational potential energy will equal the maximum kinetic energy. The maximum GPE will equal when the stone is at its highest point and the max KE will be right before the stone hits the ground. So

GPE max = KE max

mgh = 1/2mv^2

Mass cancels as it is on both sides

gh = 1/2v^2

Multiply by 2

2gh = v^2

Square root

v = √2gh

Now plug in

v = √2(9.8)(4)

v = 8.85 m/s

Now use the mass to calculate the momentum just before it hits the ground as this is the speed right before it hits the ground

p = mv

p = (3)(8.85)

p = 26.55 kgm/s

true

explanation:

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y