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 1.75 m wide and 9.00 m below the top of the cliff?

1.292 m/s

Explanation:

For her trajectory motion, the time taken to descend to a height of 9 m is given as

t = √(2H/g) = √[(2×9)/9.8] = 1.355 s

Then the range (horizontal distance covered) is related to the time to reach that height and the corresponding horizontal velocity as shown in

R = ut

where u = initial horizontal velocity

1.75 = u × 1.355

u = 1.292 m/s

This is the minimum speed just as she leaves the top of the cliff to miss the ledge described in the question.

part 1: which substance has a higher solubility?

a

smallest amount left undissolved atall temps in table shown most has dissolved high sol'by