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

Question

What must her mim speed be just as she leaves the top of the cliff so that she will miss a 1.75 m wide and 9 m ledge below the top of the cliff ?

Answer 1

Break this into horizontal and vertical components

x(t)=v0*t
y(t)=-.5*g*t^2

when y(t)=-9, x(t)>1.75
using g=9.81
t=sqrt(18/9.81) when the diver falls to the height of the ledge
1.75 solve for v0
v0>1.29 m/s

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