A tourist accidentally drops a camera from a 40.0 m high bridge. What is the speed of the camera as it hits the water?

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Disregard air resistance.

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Answer 1 · 2018-03-10T13:24:29

#28 ms^-1#

Apply law of conservation of energy to find the velocity while hitting the water.

Total energy while in the hand of the tourist is purely potential energy i.e #mgh=40mg# (where, #m# is the mass of the camera)

And total energy while hitting the water is purely kinetic energy i.e #1/2 mv^2# where, #v# is the velocity with which the camera hits water.

So,we can equate both,

#40mg=1/2mv^2#

so, #v=sqrt(2*40g)=28 ms^-1#