Question #22380

A 3.5 kg block falls from a height of 12 m. What is its momentum just before it hits the ground?

Expert's answer

The velocity of a block just before it hits the ground is

V = g√(2H/g) = 9.8[m/s²]√(2·12[m]/9.8[m/s²]) ≈ 58.7[m/s],

where √(2H/g) is the time of fall. So, the momentum of a block just before it hits the ground is

I = M·V = 3.5[kg]·58.7[m/s] = 205.45 kg·[m/s].

V = g√(2H/g) = 9.8[m/s²]√(2·12[m]/9.8[m/s²]) ≈ 58.7[m/s],

where √(2H/g) is the time of fall. So, the momentum of a block just before it hits the ground is

I = M·V = 3.5[kg]·58.7[m/s] = 205.45 kg·[m/s].

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