Answer:
The diver can jump a height of 4m in the swimming area.
Explanation:
Using the kinematics equation of the final velocity in terms of the distance, we have that:
[tex]v_{fy}^{2} =v_{oy}^{2}-2gy[/tex]
Since the final velocity is zero (because the diver comes to a stop in his maximum height), we conclude:
[tex]v_{oy1}^{2} =2g_1y_1;v_{oy2}^{2} =2g_2y_2[/tex]
If we assume that the initial velocity is conserved (as the diver jumps with this initial velocity every time), we obtain:
[tex]v_{oy1}^{2}=v_{oy2}^{2}\\\\2g_1y_1=2g_2y_2\\\\y_2=\frac{g_1y_1}{\frac{1}{4} g_1} \\\\y_2=4y_1\\\\\implies y_2=4m[/tex]
In words, the diver can jump a height of 4m in the swimming area.