V^{2}_{f}=V^{2}_{i} +2ad ;
02 = 32 + 2 (-9.8)d;
0 = 9 + (-19.6)*d; -9 =
(-19.6)d; d = 0.46 m.
Chloe will travel 0.46 meters upward due to the takeoff velocity
of 3 m/s. Or Chloe's center of gravity will be 2.96 m (2.5 + 0.46)
above the water level.
b. Chloe's time in the sir includes time to travel upward and time
to travel downward until his hands touches the water. The time
travel upward depends on the takeoff velocity and the gravity ( ).
; 0 = 3 + (-9.8)t; -3 = (-9.8)t; t = 0.31 seconds. The time travel
downward depends on the gravity and displacement. The total
vertical displacement will be 0.46 m plus 1.5 m this is due to his
final position where his center gravity is 1 meter above the water.
Therefore, the displacement from the end of the previous downward
travel is 2.5-1 = 1.5 m. So the total downward displacement is 1.96
meters. ; -1.96 = 0 t + 0.5 (-9.8)t2; -1.96 =
(-4.9) t2; t2 = 0.4; t = 0.63 seconds which is the time for
traveling downward. The total time in the air will be 0.31 + 0.63 =
0.94 seconds.
calculation may be wrong but concept is right.
Chloe has a vertical velocity of 3 m/s when she leaves the 1m diving board
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