(Un) Realistic Situation #2

Classical Mechanics Level 2

A man of mass 60 k g 60 kg having a bag of mass 2 k g 2 kg slips from the roof of a tall building of height 128 m 128 m & starts falling vertically. When at a height of 110 m 110 m from the ground, he notices that the ground below is very hard, but that there is a pond at a distance of 10 m 10m from the line of fall.

In order to save himself from the fall, he throws the bag horizontally (with respect to himself) in the direction opposite to the pond.

Calculate the minimum velocity imparted to the bag so that the man lands in the water.

You can also try (Un) Realistic Situation #1 .

44.7 m/s 93.9 m/s 4.2 m/s None of These

Ameya Salankar by Ameya Salankar , 23, India

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1 solution

Sungil Cho
May 15, 2014

The remaining time to fall from the height of 110m is

T = 2 × 128 g − 2 × 18 g = 10 g T=\sqrt{\frac{2×128}{g}}-\sqrt{\frac{2×18}{g}}=\frac{10}{\sqrt{g}}

The speed needed to reach the pond is

v = 10 T = g v=\frac{10}{T}=\sqrt{g}

From the momentum conservation the speed of the bag is

v ′ = 30 × v = 30 g v'=30×v=30\sqrt{g}

7 Helpful 0 Interesting 0 Brilliant 0 Confused

could you please explain more the third step ?

Kimo El-ghazawy - 7 years ago

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By law of conversation of momentum : mass * velocity =constant So, we have mass of bag * velocity of bag = mass of the man * velocity of the man

Tirukkovalluri Sandeep - 7 years ago

Actually the fall from 110 to the ground is not free fall it is the case of horizontal projectile then how can you subtract the time

Phani Burlu - 3 years, 2 months ago

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