Question

A daring 510N swimmersdives off a cliff with a running horizontal lead.what must be her mimimum speed just as she leaves the top of cliff so that she will miss the ledge at the bottom which is 1.75 m wide and 9m below the top of cliff

Answer 1

Answer:

The running horizontal speed should be larger than 1.29 m/s.

Explanation:

In order for the swimmer to just miss the bone-breaking ledge, her horizontal speed must be

$v > \frac{1.75m}{t_{fall}}$

in which we need to know how long we she be diving through the air. To determine that, recall the formula for the distance made by an object with acceleration (in this case it is the gravitational acceleration) with no initial (vertical) velocity:

$s = \frac{1}{2}gt^2$

from which it follows that (for non-negative t)

$t = \sqrt{\frac{2\cdot9m}{9.8\fram{m}{s^2}}}\approx1.36s$

This result can be used in the initial inequality:

$v > \frac{1.75m}{t_{fall}}=\frac{1.75m}{1.36s}=1.29\frac{m}{s}$

The diving lady better gets a speed larger than 1.29 m/s to avoid landing on the ledge.

Answer 2

looks like its projectiles = swimmer. see encls may help