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
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.
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