Respuesta :
The initial velocity is
v(0) = 16.5 ft/s
While in the water, the acceleration is
a(t) = 10 - 0.[tex] \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}[/tex]
The velocity function is
[tex]v(t)=12.5+4e^{-0.8t}[/tex]
It satisfies the condition that v(0) = 16.5 ft/s.
When t = 5.7s, obtain
[tex]v(5.7)=12.5+4e^{-0.8\times5.7} = 12.54 \, ft/s[/tex]
The depth of the lake is
[tex]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[/tex]
Answer:
The velocity at the bottom of the lake is 12.5 ft/s
The depth of the lake is 76.2 ft
v(0) = 16.5 ft/s
While in the water, the acceleration is
a(t) = 10 - 0.[tex] \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}[/tex]
The velocity function is
[tex]v(t)=12.5+4e^{-0.8t}[/tex]
It satisfies the condition that v(0) = 16.5 ft/s.
When t = 5.7s, obtain
[tex]v(5.7)=12.5+4e^{-0.8\times5.7} = 12.54 \, ft/s[/tex]
The depth of the lake is
[tex]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[/tex]
Answer:
The velocity at the bottom of the lake is 12.5 ft/s
The depth of the lake is 76.2 ft
The velocity of the ball at the bottom of the lake is 12.54 m/s.
Acceleration of the ball
The acceleration of the ball while in the water is calculated as follows;
[tex]a = \frac{dv}{dt} \\\\10-0.8v = \frac{dv}{dt} \\\\\frac{dv}{10-0.8v}= dt\\\\ \int\limits^v_{v_0} {\frac{dv}{10-0.8v} } = \int\limits^t_0 {dt} \\\\-\frac{1}{0.8} [ln(10-0.8v)]^v_{v_0} = t\\\\ln(\frac{10-0.8v}{-3.2} ) = -0.8t\\\\\frac{0.8v-10}{3.2} = -0.8t\\\\\frac{0.8v-10}{3.2} = e^{-0.8t} \\\\0.8v = 10 + 3.2e^{-0.8t}[/tex]
[tex]v = \frac{10}{0.8} + \frac{3.2e^{-0.8t}}{0.8} \\\\v (t) = 12.5 + 4e^{-0.8t}[/tex]
Velocity of the ball at the bottom of the lake
The velocity of the ball at the bottom of the lake is determined by applying the equation above at time 5.7 s.
[tex]v(5.7) = 12.5 + 4e^{-0.8\times 5.7}\\\\v(5.7) = 12.5 + 0.04\\\\v(5.7) = 12.54 \ m/s[/tex]
Thus, the velocity of the ball at the bottom of the lake is 12.54 m/s.
Learn more about velocity of a dropped object here: https://brainly.com/question/18732868
