Question

Please Help!!

During a cliff diving competition, a diver begins a dive with his center of gravity 70ft above the water. The initial vertical velocity of his dive is 8ft per second.

a. Write and graph an equation that models the height [h] (in feet) of the diver's center of gravity as a function of time [t] (in seconds).

b. How long after the diver begins his dive does his center of gravity reach the water?

Answer 1

if you take the initial velocity to be up(opposite of the ground) im saying that to clarify if im wrong. h = ho +vot +1/2at^2

Answer 2

i think you use (since you are working in feet)
\[h(t)=-16t^2+8t+70\] and so set it equal to zero to solve for second problem

Answer 3

yes :)

Answer 4

So, it takes 70 seconds?

Answer 5

First, put down everything you know:

Vi = -8 ft/s
Xi = 70ft
Xf = 0ft
a = -32ft/s^2

Now use the kinematic formulas. First, find Vf (final velocity) using \[Vf^2=Vi^2+2a(Xf-Xi)\]
So it's: \[Vf^2 = (8ft/s)^2 + 2 (-32 ft/s^2)(-70ft) = 4544 ft/s\]
\[Vf=\sqrt{4544}=67.41ft/s\]
Once you know Vf, solving for time is pretty easy using: \[Vf=Vi+at\]
So it's \[67.41ft/s = 8ft/s + (32ft/s^2)t\]
Solve for t to get: ~1.86 seconds