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remarks notice that the speed halfway down is not half the final speed. another interesting point is that the final answer doesnt depend on the mass. that is really a consequence of neglecting the change in kinetic energy of earth, which is valid when the mass of the object, the diver in this case, is much smaller than the mass of earth. in reality, earth also falls towards the diver, reducing the final speed, but the reduction is so minuscule it could never be measured.
question qualitatively, how will the answers change if the diver takes a running dive off the end of the board?
answer (a) would increase and answer (b) would decrease.
answer (b) would increase and answer (a) would decrease.
both answers would decrease.
both answers would increase.
both answers would remain the same.
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use the worked example above to help you solve this problem. a diver of mass m drops from a board 16.0 m above the waters surface, as shown in the figure. neglect air resistance.
(a) use conservation of mechanical energy to find his speed 8.00 m above the waters surface. 12.5 m/s
(b) find his speed as he hits the water. 17.7 m/s

Explanation:

Step1: Analyze energy - initial condition change

When the diver takes a running dive, the initial kinetic energy is non - zero.

Step2: Consider conservation of mechanical energy

$E = K+U$. For part (a), at 8.00 m above water, with more initial $K$, the total mechanical energy is higher. Using $mgh_1+\frac{1}{2}mv_{0}^2=mgh_2+\frac{1}{2}mv_{2}^2$, with larger initial $K$, the speed at 8.00 m above water ($v_2$) will increase. For part (b), when hitting the water, again with more initial $K$, using $mgh_1+\frac{1}{2}mv_{0}^2=\frac{1}{2}mv_{f}^2$, the final speed $v_f$ will increase.

Answer:

Both answers would increase.