An Olympic diver drops from the 10 meter platform with an initial velocity of 0.0m/s. What was the vertical velocity of the diver immediately before he/she hits the water? How long did it take the diver to reach the water?

v=14.14 m/s

t=1.141 s    

Explanation:

Given that

h = 10 m

Initial velocity ,u = 0 m/s

We know that acceleration due to gravity g= 10 m/s²

Lets take final velocity = v

Final velocity v is given as

v² = u²+ 2 g h

v²= 0² + 2 x 10 x 10

v²= 200

v=14.14 m/s

Time taken t is given as

v= u +  a t

a=acceleration

t=time

Now by putting the values in the above equation we get

14.14= 0 + 10 x t

14.14 = 10 t

t=1.141 s

i think if it were me i whould divide 300 dide 15

step-by-step explanation:

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s