a diver dives from a cliff .when her centre of gravity is 46 ft above the surface of the water. her initial vertical velocity leaving the water is 9 ft per second. after how many secs does she hit the water? solve

5 answers

At what point in the dive does she "leave the water" ? I can't picture what is going on. Does she dive initially vertically upward?
You must use the equation h=-16t^2+vt+s

h=0
v=9
s=46

Then plug in and solve (I plugged the numbers into the equation but I'll let you solve it!) Hope this helps!

0=-16t^2+9t+46
Use the equation h=-16^2+vt+s

Plug in the numbers. In this case,

h=0
v=9
s=46

You will get 0=-16t^2+9t+46
(a) (b)(c)

Then use the quadratic formula and plug in the numbers to that.
1.475
So add 6 plus 09 and add -93