Differential Equations as Mathematical Models-Falling bodies
Posted: Tue May 12, 2015 6:11 pm
In this topic, we formulate and solve a number of differential equations that originate from real-world problems.
Example 1
A ball of mass is thrown straight up from the ground with an initial velocity of . Ignoring air resistance, find the maximum height and the time the ball takes to attain the maximum height.
Solution
Let the height of the ball from the ground at any time be .
Since the force acting on the ball is , Newton's second law of motion, , gives the relation
Thus
Integrating both sides,
\begin{equation}\tag{eq1}
m\dfrac{dy}{dt} + gt + c = 0
\end{equation}
Velocity at , so (eq1) yields
, this
Now, we have
\begin{equation}\tag{eq2} \dfrac{dy}{dt} + gt - v_{0} = 0 \end{equation}
Integrating (eq2),
Since
The equation of position at any time is
\begin{equation}\tag{eq3} y(t) = -\frac{1}{2}gt^{2} + v_{0}t \end{equation}
To find time () at the maximum height, we set the first derivative of (eq3) equals to (set velocity = ) to have
, .
Check that the second derivative at this point is negative, i.e., .
Maximum height possible the ball can attain is
.
Example 2
A ball weighing is thrown vertically upwards from a point above the surface of the earth with an initial velocity of . As it rises it is acted upon by air resistance that is numerically equal to (in pounds), where is the velocity in feet per second. How high will the ball rise?
Solution
Let the height of the ball at at any time from the ground be , and assume that the upward direction is positive. From Newton's second law of motion , where is the mass of an object and is acceleration due to gravity,
,
Separating variables,
Integrating,
\begin{equation}\tag{1}-\frac{3}{2}ln(v+48) = t + c_{1} \end{equation}
Applying ,
From
\end{equation}\tag{2} v = e^-\left({\dfrac{2t + c_{2}}{3}}\right) - 48[/tex], where c_2 = -3ln(68) \end{equation}
At the maximum height, ,
So , where is the time the ball takes to reach maximum height.
Hence \begin{equation}\tag{3} y(t)_{max} = \int_{0}^{t_{max}}\left( e^-\left({\dfrac{2t + c_{2}}{3}}\right) - 48 \right)dt \ + 6 \end{equation}
To be continued.
Example 1
A ball of mass is thrown straight up from the ground with an initial velocity of . Ignoring air resistance, find the maximum height and the time the ball takes to attain the maximum height.
Solution
Let the height of the ball from the ground at any time be .
Since the force acting on the ball is , Newton's second law of motion, , gives the relation
Thus
Integrating both sides,
\begin{equation}\tag{eq1}
m\dfrac{dy}{dt} + gt + c = 0
\end{equation}
Velocity at , so (eq1) yields
, this
Now, we have
\begin{equation}\tag{eq2} \dfrac{dy}{dt} + gt - v_{0} = 0 \end{equation}
Integrating (eq2),
Since
The equation of position at any time is
\begin{equation}\tag{eq3} y(t) = -\frac{1}{2}gt^{2} + v_{0}t \end{equation}
To find time () at the maximum height, we set the first derivative of (eq3) equals to (set velocity = ) to have
, .
Check that the second derivative at this point is negative, i.e., .
Maximum height possible the ball can attain is
.
Example 2
A ball weighing is thrown vertically upwards from a point above the surface of the earth with an initial velocity of . As it rises it is acted upon by air resistance that is numerically equal to (in pounds), where is the velocity in feet per second. How high will the ball rise?
Solution
Let the height of the ball at at any time from the ground be , and assume that the upward direction is positive. From Newton's second law of motion , where is the mass of an object and is acceleration due to gravity,
,
Separating variables,
Integrating,
\begin{equation}\tag{1}-\frac{3}{2}ln(v+48) = t + c_{1} \end{equation}
Applying ,
From
\end{equation}\tag{2} v = e^-\left({\dfrac{2t + c_{2}}{3}}\right) - 48[/tex], where c_2 = -3ln(68) \end{equation}
At the maximum height, ,
So , where is the time the ball takes to reach maximum height.
Hence \begin{equation}\tag{3} y(t)_{max} = \int_{0}^{t_{max}}\left( e^-\left({\dfrac{2t + c_{2}}{3}}\right) - 48 \right)dt \ + 6 \end{equation}
To be continued.