# 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 $$127.551{\text{m}}$$.