Question

Since raindrops grow as they fall, their surface area increases, and therefore the resistance to their falling increase. A raindrop has an initial downward velocity of 10 m/s and its downward acceleration is

\(a = \left\{ \begin{aligned}{l}9 - 0.9t{\rm{ 0}} \le {\rm{t}} \le {\rm{10}}\\{\rm{0 t > 0}}\end{aligned} \right.\)

If the raindrop is initially 500 m above the ground. How long does it take to fall?

Short answer

It takes \(11.82\) seconds for the raindrop to hit the ground.

Step-by-step solution

Given information

In the question given downward acceleration is,

\(a = \left\{ \begin{aligned}{l}9 - 0.9t{\rm{ 0}} \le {\rm{t}} \le {\rm{10}}\\{\rm{0 t > 0}}\end{aligned} \right.\)

Now we can make it upward direction,

\(a = \left\{ \begin{aligned}{l} - 9 + 0.9t{\rm{ 0}} \le {\rm{t}} \le {\rm{10}}\\{\rm{0 t > 0}}\end{aligned} \right.\)

Velocity is the antiderivative of acceleration .so,

Let \(f\left( x \right) = - 9 + 0.9t\).

Find the derivative

We find it derivative,

\(\begin{aligned}{c}f'\left( x \right) &= - 9t + 0.9\frac{{{t^2}}}{2} + {c_1}\\ &= - 9t + 4.5{t^2} + {c_1}\end{aligned}\)

So,

\(a = \left\{ \begin{aligned}{l}{\rm{ }} - 9t + 4.5{t^2} + {c_1}{\rm{ 0}} \le {\rm{t}} \le {\rm{10}}\\{\rm{0 t > 0}}\end{aligned} \right.\)

It is given that \({v_0} = - 10\frac{m}{s}\).

\({v_0} = - 10 = {c_1}\)

\(a = \left\{ \begin{aligned}{l}{\rm{ }} - 9t + 4.5{t^2} + 10{\rm{ 0}} \le {\rm{t}} \le {\rm{10}}\\{\rm{0 t > 0}}\end{aligned} \right.\)

Because in the question velocity is continuous so \({c_2}\)should be the same as \(v\left( {10} \right)\).

\(\begin{aligned}{c}{c_2} &= - t + 0.45{t^2}\\ &= v\left( {10} \right)\\ &= - 9t + 4.5{t^2} + {c_1}\end{aligned}\)

Find the time

Find the constant by putting \(t = 10\).

\(\begin{aligned}{c}{c_2} &= - 9\left( {10} \right) + 4.5{\left( {10} \right)^2} - 10\\ &= - 9\left( {10} \right) + 4.5{\left( {10} \right)^2} - 10\\ &= - 90 + 45 - 10\\ &= - 100 + 45\\ &= - 55\end{aligned}\)

Then,

\(a = \left\{ \begin{aligned}{l}{\rm{ }} - 9t + 4.5{t^2} + 10{\rm{ 0}} \le {\rm{t}} \le {\rm{10}}\\{\rm{ - 55 t > 0}}\end{aligned} \right.\)

Position is the \({s_0} = 500\).

Here we differentiate v.

\(\begin{aligned}{c}{v^1} &= - 9t + 4.5{t^2} + 10\\v'' &= - 9\frac{{{t^2}}}{2} + 0.45\frac{{{t^3}}}{3} - 10t + {s_0}\\ &= - 4.5{t^2} + 0.15{t^3} - 10t + 500\end{aligned}\)

Let \(u = - 55\)

\(u' = - 55t + {c_3}\)

We arrange it in this form,

\(a = \left\{ \begin{aligned}{l}{\rm{ }} - 4.5{t^2} + 0.15{t^3} - 10t + 500{\rm{ 0}} \le {\rm{t}} \le {\rm{10}}\\{\rm{ }} - 55t + {c_3}{\rm{ t > 0}}\end{aligned} \right.\)

After 10 seconds, the raindrop position is,

\(\begin{aligned}{c}s\left( {10} \right) &= - 4.5{\left( {10} \right)^2} + 0.15{\left( {10} \right)^3} - 10\left( {10} \right) + 500\\ &= - 4.5 \times 100 + 0.15 \times 1000 - 100 + 500\\ &= 100m\end{aligned}\)

Raindrops fall the last \(100m\) when the velocity is \(55\frac{m}{s}\).

\(\begin{aligned}{c}100 &= 55t\\t &= \frac{{100}}{{55}}\\ &\approx 1.82\end{aligned}\)

It takes 10 seconds to fall 400m, then about 1.82 seconds to fall the last 100 m.

So, it takes 11.82 seconds for the raindrop to hit the ground.