a) Assume the acceleration of gravity is 9.8m/s² we can write:
-1.30=3t-4.9t², (upward distance resulting from her take-off speed, less the downward distance fallen under gravity) where t is the time it takes for her feet to reach the water 1.30m below the diving board. This can be written: 4.9t²-3t-1.30=0. This quadratic equation can be solved by completing the square:
t²-(3/4.9)t=1.30/4.9,
t²-(3/4.9)t+(3/9.8)²=1.30/4.9+(3/9.8)²=0.3590 approx.
(t-3/9.8)²=0.3590, t=3/9.8+√0.3590=0.9053 secs approx. or 0.91 rounded to 2 decimal places.
b) In general terms, the height h above the diving board is given by:
h=3t-4.9t². This can also be written:
h=4.9(3t/4.9-t²)=4.9((3/9.8)²-(3/9.8)²+3t/4.9-t²)
h=4.9(3/9.8)²-4.9(t-3/9.8)².
From this it’s easy to see that the maximum height occurs when t=3/9.8 secs because 4.9(t-3/9.8)² is zero so nothing is subtracted from 4.9(3/9.8)². (All other values of t reduce this maximum.) This gives us a maximum height above the diving board of 0.4592m approx. or 0.46m rounded to two decimal places. At this maximum height her velocity becomes zero as she starts to fall towards the water.
c) Her velocity is given by v=3-9.8t (note that v=0 when t=3/9.8 secs—see also above, last sentence), t=0.9053, so v=-5.8719 m/s. Remember that the frame of reference is the diving board and positive direction is upwards from the diving board, so -5.8719 is 5.8719 in a downward direction into the water. This rounds to 5.87m/s.