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narik

Joined: 08 Nov 2009
Posts: 1

 Posted: Sun Nov 08, 2009 11:11 pm    Post subject: Kinematics Question A rock is dropped from a seacliff and the sound of it striking the ocean is heard 3.0 s later. If the speed of sound is 340 m/s, how high is the cliff? I know that t1+t2=3, but I don't know the rest.
hepcj

Joined: 23 Jun 2007
Posts: 125

 Posted: Wed Nov 18, 2009 11:17 pm    Post subject: The time is known and consists of the time it takes for the rock to fall and the time it take for the sound to reach the observer. ie. t1 + t2 = 3s. We know that the rock starts from zero velocity and it is accelerated under gravity. i.e u=0 and a =-9.81 ms^-2. From the equations of linear motion, s = ut + 1/2 at^2 or s = (1/2) a t^2 since u = 0. Rearranging, t = (2s/a)^(1/2) t is our t1. Therefore, (2s/a)^(1/2) + t2 = 3 The time t2 taken for the sound to reach the ear is the distance falllen s /(340 ms^-1) Therefore, (2s/a)^(1/2) + s/340 = 3 It is then simple to substitute the numbers and Bob's your uncle you get the height of the cliff, s.
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