A daring 510-N swimmer dives off a cliff with a running horizontal leap, as shown in ?Fig. E3.10.? What must her mini-mum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, which is 1.75 m wide and 9.00 m below the top of the cliff?

Solution 10E Step 1 of 4: First we need to find, time taken to fall past the ledge to calculate the horizontal velocity to miss the ledge. To calculate the the time taken, As there are no forces acting in the horizontal direction, there will be no horizontal acceleration (F=ma) and hence the horizontal component of velocity will be constant By fundamental relation, Velocity= Distance Time Step 2 of 4: By using equation of kinematics to calculate the time taken, S = ut + at 2 2 Since she is jumping, initial velocity will be zero, u= 0 S = at 2 2 Solving for time, 2S t = a Where S is the distance, a is acceleration, u is initial velocity and t is time.