What must her minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, which is w = 1.75 m wide and h = 8.00 m below the top of the cliff?

1.37 m/s Assuming her initial velocity is totally horizontal and her vertical velocity is only affected by gravity, let's first calculate how much time she has until she reaches the ledge 8.00 m below her. d = 1/2AT^2 8.00m = 1/2 * 9.8 m/s^2 * T^2 Solve for T 8.00 m = 4.9 m/s^2 * T^2 Divide both sides by 4.9 m/s^2 1.632653061 s^2 = T^2 Take square root of both sides 1.277753 s = T So we now know that she has 1.277753 seconds in which to reach a horizontal distance of 1.75 m. So how fast does she need to be going? 1.75 m / 1.277753 s = 1.369592 m/s Since we only have 3 significant figures in our data, round the result to 3 figures giving 1.37 m/s