Visitors at an amusement park watch divers step off a platform 21.3 (70 ) above

Question

Visitors at an amusement park watch divers step off a platform 21.3 (70 ) above a pool of water. According to the announcer, the divers enter the water at a speed of 56 ( 25 ). Air resistance may be ignored.

A) Is the announcer correct in his claim? Yes or No
B)Is it possible for a diver to leap directly upward off the board so that, missing the board on the way down, she enters the water at 25.0 ? If so, what initial upward speed is required?
C)Is the required initial speed physically attainable? Yes or No

Explanation / Answer

First of all 21.3meters=70ft.then WH=1/2mV^2=mgH=m(9.8)H,=====> 1/2V^2=21.3(9.8),==>V^2=417.48,V=20.43… enter water.But they DIVE means jump a little upwads then down,So,gH' =1/2(25)^2 H' =31.887,Dh=31.887-21.3=10.587meters above the platform again 2g(Dh)=v^2=207.5=====>v=14.4m/sec.,

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