A diver of mass m drops from a board 12.0 m above the water's surface, as shown in the figure. Neglect air resistance. Suppose the diver vaults off the springboard, leaving it with an initial speed of 3.96 m/s upward. Use energy conservation to find his speed when he strikes the water.

Formula:(1/2)(m)(Vi)+(m)(g)(Yi)=(1/2)(Vf^2)+(m)(g)(Yf)

My work: Vf=sqrt(2((1/2)(3.96)+(9.8)(3.96)-12)=15.1

Well, it seems that you've got the formula right, but there's a small mistake in your calculation. Let me help you out with the correct answer.

Using the formula (1/2)(m)(Vi) + (m)(g)(Yi) = (1/2)(Vf^2) + (m)(g)(Yf), we can plug in the known values:

Vi = 3.96 m/s (initial velocity)
Yi = 12.0 m (initial height)
Yf = 0 m (final height, at the water's surface)

Now let's solve it step by step:

(1/2)(m)(3.96) + (m)(9.8)(12) = (1/2)(Vf^2) + (m)(9.8)(0)

1.98m + 117.6m = (1/2)(Vf^2)

119.58m = (1/2)(Vf^2)

239.16m = Vf^2

Vf ≈ √(239.16m)

So the speed when the diver strikes the water would be approximately the square root of 239.16 times m, whatever the mass of the diver is.

Now here comes the fun part... I could tell you the exact speed if I knew the mass of the diver, but since I don't, let's just imagine the diver is made of marshmallows. In that case, the speed when the marshmallow diver strikes the water would be approximately √(239.16m) marshmallows per second! Just remember, that's not a very scientific unit of measurement, but it sure is tasty!

To find the speed of the diver when he strikes the water, we can use the principle of energy conservation. The formula you provided is correct:

(1/2)(m)(Vi) + (m)(g)(Yi) = (1/2)(Vf^2) + (m)(g)(Yf)

Where:
m = mass of the diver
Vi = initial speed of the diver
g = acceleration due to gravity (approximately 9.8 m/s^2)
Yi = initial height of the diver (12.0 m)
Vf = final speed of the diver (to be determined)
Yf = final height of the diver (water's surface, which we can assume to be 0 m)

Plugging in the given values:
(1/2)(m)(3.96) + (m)(9.8)(12.0) = (1/2)(Vf^2) + (m)(9.8)(0)

Simplifying the equation:
(1.98m) + (117.6m) = (1/2)(Vf^2)
119.58m = (1/2)(Vf^2)

Now, let's solve for Vf by isolating it:
Vf^2 = 2(119.58m)
Vf^2 = 239.16m
Vf = sqrt(239.16m)

Therefore, the speed of the diver when he strikes the water is sqrt(239.16m), or approximately 15.1 sqrt(m) meters per second.

To find the diver's speed when he strikes the water using energy conservation, you can use the formula you provided:

(1/2)(m)(Vi) + (m)(g)(Yi) = (1/2)(Vf^2) + (m)(g)(Yf)

Where:
m = mass of the diver
Vi = initial speed of the diver upward (3.96 m/s)
g = acceleration due to gravity (9.8 m/s^2)
Yi = initial height of the diver (12.0 m)
Vf = final speed of the diver when he strikes the water
Yf = final height of the diver (0 m, as he strikes the water)

Now let's substitute the known values into the equation:

(1/2)(m)(3.96) + (m)(9.8)(12.0) = (1/2)(Vf^2) + (m)(9.8)(0)

Simplifying the equation, the term (m)(9.8)(0) disappears:

(1/2)(m)(3.96) + (m)(9.8)(12.0) = (1/2)(Vf^2)

Now, let's solve for Vf by rearranging the equation:

(1/2)(Vf^2) = (1/2)(m)(3.96) + (m)(9.8)(12.0)

Multiply both sides by 2 to cancel out the (1/2) term:

Vf^2 = (m)(3.96) + (m)(9.8)(12.0)

Now, take the square root of both sides to isolate Vf:

Vf = √((m)(3.96) + (m)(9.8)(12.0))

This equation gives you the final speed Vf when the diver strikes the water, using energy conservation. You can substitute the specific values for mass (m) to get the numerical answer.