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A swimmer dived off a cliff with a running horizontal leap. What must his minimum speed be just as he leaves the top of the cliff so that he will miss the ledge at the bottom which is 2 m wide and 9 m below the top of the cliff?
1 Answerstime taken by swimmer to reach the ground is t
s=uyt+at^2/2 (uy is verticle component of velocity of swimmer which is zero)
s=9 and a=g=10
so t=square root(2s/g)
=3/root5 secs
horizontal distance covered during this time is t.v (v is the velocity in horizontal direction)
horizontal distance should be 2m so that he will mis the ledge
vt=2
v=2/t
=2.root5/3 m/s
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