# The Application of Differential Equations in Physics

Differential equations are commonly used in physics problems. In the following example we shall discuss a very simple application of the ordinary differential equation in physics.

Example:
A ball is thrown vertically upward with a velocity of 50m/sec. Ignoring air resistance, find

(i) The velocity of the ball at any time $t$
(ii) The distance traveled at any time $t$
(iii) The maximum height attained by the ball

Let $v$ and $h$ be the velocity and height of the ball at any time $t$. Since the time rate of velocity is acceleration, so $\frac{{dv}}{{dt}}$ is the acceleration. Since the ball is thrown upwards, its acceleration is $– g$. Thus, we have
$\frac{{dv}}{{dt}} = – g\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right)$

Separating the variables, we have
$dv = – gdt\,\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right)$

(i) Since the initial velocity is 50m/sec, to get the velocity at any time $t$, we have to integrate the left side (ii) from 50 to $v$ and its right side is integrated from 0 to $t$ as follows:

$\begin{gathered} \int_{50}^v {dv = – g\int_0^t {dt} } \\ \Rightarrow \left| v \right|_{50}^v = – g\left| t \right|_0^t \\ \Rightarrow v – 50 = – g\left( {t – 0} \right) \\ \Rightarrow v = 50 – gt\,\,\,\,{\text{ – – – }}\left( {{\text{iii}}} \right) \\ \end{gathered}$

Since $g = 9.8m/{s^2}$, putting this value in (iii), we have
$v = 50 – 9.8t\,\,\,\,{\text{ – – – }}\left( {{\text{iv}}} \right)$

(ii) Since the velocity is the time rate of distance, then $v = \frac{{dh}}{{dt}}$. Putting this value in (iv), we have
$\frac{{dh}}{{dt}} = 50 – 9.8t\,\,\,\,{\text{ – – – }}\left( {\text{v}} \right)$
Separating the variables of (v), we have
$dh = \left( {50 – 9.8t} \right)dt\,\,\,\,\,{\text{ – – – }}\left( {{\text{vi}}} \right)$

In order to find the distance traveled at any time $t$, we integrate the left side of (vi) from 0 to $h$ and its right side is integrated from 0 to $t$ as follows:

$\begin{gathered} \int_0^h {dh} = \int_0^t {\left( {50 – 9.8t} \right)dt} \\ \Rightarrow \left| h \right|_0^h = \left| {50t – 9.8\frac{{{t^2}}}{2}} \right|_0^t \\ \Rightarrow h – 0 = 50t – 9.8\frac{{{t^2}}}{2} – 0 \\ \Rightarrow h = 50t – 4.9{t^2}\,\,\,\,\,{\text{ – – – }}\left( {{\text{vii}}} \right) \\ \end{gathered}$

(iii) Since the velocity is zero at maximum height, we put $v = 0$ in (iv)
$\begin{gathered} 0 = 50t – 9.8{t^2} \Rightarrow 0 = 50 – 9.8t \\ \Rightarrow t = \frac{{50}}{{9.8}} = 5.1 \\ \end{gathered}$

Thus, the maximum height is attained at time $t = 5.1\,\sec$.

Putting this value of $t$ in equation (vii), we have
$\begin{gathered} h = 50\left( {5.1} \right) – 4.9{\left( {5.1} \right)^2} \\ \Rightarrow h = 255 – 127.449 = 127.551 \\ \end{gathered}$

Thus the maximum height attained is $127.551{\text{m}}$.