A cliff diver plunges from a height of $81 \mathrm{ft}$ above the water surface. The distance the diver falls in $t$ seconds is given by the function $d(t)=16 t^{2} \mathrm{ft}$. (a) After how many seconds will the diver hit the water? s (b) With what velocity (in $\mathrm{ft} / \mathrm{s}$ ) does the diver hit the water? $\mathrm{ft} / \mathrm{s}$

Step 1 ：The distance the diver falls in t seconds is given by the function \(d(t)=16 t^{2} \mathrm{ft}\).

Step 2 ：We can find when the diver will hit the water by setting the distance function equal to the height of the cliff and solving for time. So, we set \(d(t) = 81\) and solve for \(t\).

Step 3 ：Solving the equation \(16t^2 = 81\) gives us two solutions for \(t\), which are \(-\frac{9}{4}\) and \(\frac{9}{4}\). However, time cannot be negative, so we discard the negative solution.

Step 4 ：Therefore, the diver will hit the water after \(\boxed{\frac{9}{4}}\) seconds.

Step 5 ：We can find the velocity of the diver when they hit the water by taking the derivative of the distance function, which gives us the velocity function \(v(t) = 32t\).

Step 6 ：Evaluating the velocity function at the time the diver hits the water, \(t = \frac{9}{4}\), gives us a velocity of \(72 \, \mathrm{ft/s}\).

Step 7 ：Therefore, the diver will hit the water with a velocity of \(\boxed{72 \, \mathrm{ft/s}}\).