Question

It is intuitive that if you are jumping from a high cliff, you would rather land in water than on sand -- but why?

Group of answer choices

a) In bringing you to a halt, the sand exerts a greater impulse on you than does the water

b) In bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a greater average force.

c) In bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a smaller average force.

Answer 1

We know that Impluse is given by:

Impulse = change in momentum

J = dP = m*dV = m*(Vf - Vi)

Now when you'are jumping from a high cliff, than

Vi = Initial velocity just before hitting sand/water (It will be same in both case since initial height is same)

Vf = final Velocity = 0 m/s, since finally you'are halted by water/sand

m = mass of you (Again same in both scenarios)

So,

J = m*(Vf - Vi)

Impulse exerted by both water and sand will be same.

Now average force applied on you be water or sand will be:

F_avg = dP/dt

dt = impact time = time taken by you to stop

Now when we land on water than impact time (stopping time) is higher, and since average force is inversely proportional to impact time, So higher time means average force applied by water will be lower.

And when we land on sand than impact time (stopping time) is lower, and since average force is inversely proportional to impact time, So lower time means average force applied by sand will be higher.

So In bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a greater average force.

Correct option is B.

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