想問下Physics 09年CE Question 28 條點計，係睇gragh 既，upload 唔到圖，希望有人教?

The total distance travelled by the diver is the area bounded by the straight lines and the x-axis.

From 0 to 0.3 s, area of triangle
= (1/2) x 0.3 x 3 m = 0.45 m
From 0.3 s to 2 s, area of triangle
= (1/2) x (2 - 0.3) x 13 m = 11.05 m
Hence, total distance travelled
= (0.45 + 11.05) m = 11.5 m

Consider the path from P to water surface. The diver is under free fall.
Use equation of motion: v^2 = u^2 + 2as
with u = 3 m/s, v = -13 m/s, a = -g(= 10 m/s^2), s =?
Hence, (-13)^2 = 3^2 + 2.(-10).s
i.e. s = (169 - 9)/(-20) m = -8 m
(the -ve sign indicates that the water surface is 8 m BELOW the platform.)

Therefore, height of platform = 8 m
The answer is option C