Question

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 the water, the ball experiences an acceleration of a = 10 – 0.8v, where a and v are expressed in ft/s2 and ft/s, respectively. The ball takes 5.7 s to reach the bottom of the lake.

Answer 1

The initial velocity is
v(0) = 16.5 ft/s

While in the water, the acceleration is
a(t) = 10 - 0.$\frac{dv}{dt} =10-0.8v \\\\ \frac{dv}{10-0.8v}=dt \\\\ \int_{16.5}^{v} \, \frac{dv}{10-0.8v} = \int_{0}^{t} dt \\\\ - \frac{1}{0.8} [ln(10-0.8v)]_{16.5}^{v}=t \\\\ ln \frac{10-0.8v}{-3.2}=-0.8t \\\\ \frac{0.8v -10}{3.2} =e^{-0.8t} \\\\ 0.8v = 10 + 3.2e^{-0.8t} \\\\ v=12.5+4e^{-0.08t}$

The velocity function is
$v(t)=12.5+4e^{-0.8t}$
It satisfies the condition that v(0) = 16.5 ft/s.
When t = 5.7s, obtain
$v(5.7)=12.5+4e^{-0.8\times5.7} = 12.54 \, ft/s$

The depth of the lake is
$d=\int_{0}^{5.7} \, (12.5+4e^{-0.8t})dt \\\\ = 12.5(5.7)+ \frac{4}{(-0.8)}[e^{-0.8t}]_{0}^{5.7} \\\\ =71.25-5(0.0105-1) =76.198 \, ft$

Answer:
The velocity at the bottom of the lake is 12.5 ft/s
The depth of the lake is 76.2 ft