a ball is dropped from a boat so that it strikes the surface of a lake with a speed of 16.5 ft/s. while in…

Explanation:

Step1: Recall the kinematic - equation for acceleration

The kinematic equation $v = v_0+at$ and $x = v_0t+\frac{1}{2}at^{2}$ can be used. Here, the initial velocity $v_0 = 16.5$ ft/s, $a=10 - 0.8v$, and $t = 3$s.

Step2: First, use the velocity - time relation

We know that $v=v_0+\int_{0}^{t}a\mathrm{d}t$. Substituting $a = 10 - 0.8v$ into the equation, we get $\frac{\mathrm{d}v}{10 - 0.8v}=\mathrm{d}t$. Integrating both sides: [-\frac{1}{0.8}\ln|10 - 0.8v|=t + C] When $t = 0$, $v = v_0=16.5$ ft/s. [-\frac{1}{0.8}\ln|10 - 0.8\times16.5|=C] [-\frac{1}{0.8}\ln|10 - 13.2|=C] [-\frac{1}{0.8}\ln(3.2)=C] After integrating $\int_{16.5}^{v}\frac{\mathrm{d}v}{10 - 0.8v}=\int_{0}^{3}\mathrm{d}t$: [-\frac{1}{0.8}[\ln|10 - 0.8v|-\ln|10 - 0.8\times16.5|]=3] [-\frac{1}{0.8}\ln\left|\frac{10 - 0.8v}{10 - 13.2}\right|=3] [\ln\left|\frac{10 - 0.8v}{- 3.2}\right|=-2.4] [\frac{10 - 0.8v}{-3.2}=e^{-2.4}] [10 - 0.8v=-3.2e^{-2.4}] [0.8v = 10+3.2e^{-2.4}] [v=\frac{10 + 3.2e^{-2.4}}{0.8}] [v=\frac{10+3.2\times0.0907}{0.8}=\frac{10 + 0.29024}{0.8}=\frac{10.29024}{0.8}=12.8628\approx12.9] ft/s.

Step3: Use the position - time relation

We also know that $x=v_0t+\frac{1}{2}\int_{0}^{t}a\mathrm{d}t\times t$. Another way is to use the non - linear differential equation approach. But we can also use the fact that we first found the final velocity $v$. We can use the average - velocity formula $x=\frac{v_0 + v}{2}t$ (since the acceleration is non - constant, this is an approximation, a more accurate way is to use $x=\int_{0}^{t}(v_0+\int_{0}^{t}a\mathrm{d}t)\mathrm{d}t$). [x=\int_{0}^{3}(16.5+(10 - 0.8v)t)\mathrm{d}t] First, from $v = v_0+\int_{0}^{t}a\mathrm{d}t$, we can also use the fact that we know $v$ at $t = 3$s. Using the more accurate integral method: [x=\int_{0}^{3}\left(16.5+(10 - 0.8\left(16.5+\int_{0}^{t}(10 - 0.8v)\mathrm{d}t\right))t\right)\mathrm{d}t] A simpler way is to use the kinematic equation $x=v_0t+\frac{1}{2}at^{2}$ in a non - standard way. We know that $a = 10-0.8v$. We can approximate the depth $x$ as follows: [x=16.5\times3+\frac{1}{2}\left((10 - 0.8\times16.5)+(10 - 0.8\times12.9)\right)\times3\times\frac{1}{2}] [x = 49.5+\frac{3}{4}(10 - 13.2+10 - 10.32)] [x = 49.5+\frac{3}{4}( - 3.2 - 0.32)] [x = 49.5+\frac{3}{4}\times(-3.52)] [x = 49.5-2.64=46.86\approx46.9] ft.

(a)

Answer:

The depth of the lake is approximately $46.9$ ft. (b)

Answer:

The speed of the ball when it hits the bottom of the lake is approximately $12.9$ ft/s.