Adiver running 1.8 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 3s later. how high was the cliff and how far from its base did the diver hit the water? work needed

We use kinematic equation,

Here, h is vertical height, u is initial vertical velocity, t is time taken and g is acceleration due to gravity.

As diver dives out horizontally, his velocity is directed horizontally; that is, the initial vertical velocity is 0. So above equation becomes

Given, t =3 s.

Therefore,

.

Now the horizontal distance of the diver to hit the water from base,  

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

v patient student

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form