A 510 N swimmer dives off cliff with a horizontal leap...?

Question

what is the minimum speed to leap off the cliff to miss the ledge at the bottom that is 1.75 m wide and 9.00 m below the cliff?

v= m/s

Answer 1

When the diver is at y(t)=-9, x(t) must be greater than 1.75 m. The minimum vx is when x(t)=1.75,
so
1.75=v*t
and
-9=-.5*g*t^2
t=sqrt(18/g)
v=1.75/Sqrt(18/g)

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