Lecture 11 Cha nbox (9431-b LEARN MORE REMARKS Notice that the speed halftway do

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Lecture 11 Cha nbox (9431-b LEARN MORE REMARKS Notice that the speed halftway down is not half the final speed. Another interesting point is that the final answer doesn't depend on the mass. That is rally 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, bu the reduction is so minuscule it could never be measured. QUESTION Qualiltatively, how will the answers change if the diver takes a running dive off the end of the board? O Answer (a) would increase and answer (b) would decrease. O Answer (b) would increase and answer (a) would decrease. O Both answers would decrease. O Both answers would increase O Both answers would remain the same. PRACTICE IT Use the worked example above to heip you solve this problem. A diver of mass m drops from a board 6.00 m above the water's surface, as shown in the figure. Neglect air resistance. (a) Use conservation of mechanical energy to find his speed 3.00 m above the water's surface. m/s (b) Find his speed as he hits the water m/s EXERCISE HINTS: GETTING STARTED rM STUCKI Use the values from PRACTICE IT to help you work this exercise. Suppose the diver vaults off the springboard, leaving it with an initial speed of 3.09 m/s upward. Use energy conservation to find his speed when he strikes the water m/s Need Help? Talk to a Tuter

Explanation / Answer

practice problems]

a] by energy conservstion, 0.5 mv^2 = mgh where h is change in height

v = sqrt(2gh) = sqrt(2*9.8*3) = 7.67 m/s

b] again v = sqrt(2*9.8*6) = 10.844 m/s

exercise] By energy conservation : 0.5 mv^2 = 0.5 mu^2 + mgh

v = sqrt(u^2+2gh) = sqrt(3.09^2+2*9.8*6) = 11.276 m/s

Question] both would increase as it will have an additional KE

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