To solve this problem, we will use the kinematic equations for uniformly accelerated motion.$\\$Given:$\\$• Initial velocity $u = 4$ m/s (upwards).$\\$• Time of flight $t = 4$ s.$\\$• Acceleration due to gravity $g = 10$ m/s$^2$ (downwards).$\\$We need to find the height of the bridge above the water surface.$\\$Let's denote:$\\$• $h$ as the height of the bridge.$\\$• $s_1$ as the maximum height reached above the bridge.$\\$• $s_2$ as the distance the ball falls from the maximum height to the water surface.$\\$First, find the time taken to reach the maximum height:$\\ v = u - gt \\ 0 = 4 - 10t_1 \\ t_1 = \frac{4}{10} = 0.4$ s$\\$Now, calculate the maximum height $s_1$ above the bridge:$\\ s_1 = ut_1 - \frac{1}{2}gt_1^2 \\ s_1 = 4 \times 0.4 - \frac{1}{2} \times 10 \times (0.4)^2 \\ s_1 = 1.6 - 0.8 = 0.8$ m$\\$The total time of flight is 4 s, so the time taken to fall from the maximum height to the water surface is:$\\ t_2 = 4 - t_1 = 4 - 0.4 = 3.6$ s$\\$Calculate the distance $s_2$ the ball falls from the maximum height to the water surface:$\\ s_2 = \frac{1}{2}gt_2^2 \\ s_2 = \frac{1}{2} \times 10 \times (3.6)^2 \\ s_2 = 5 \times 12.96 = 64.8$ m$\\$The height of the bridge $h$ is the difference between $s_2$ and $s_1: \\ h = s_2 - s_1 \\ h = 64.8 - 0.8 = 64$ m$\\$Therefore, the height of the bridge above the water surface is $64$ m.$\\$This corresponds to Option 3.