Question

A projectile leaves the ground with an initial speed of \[20{\rm{ m}}{{\rm{s}}^{ - 1}}\] at an angle \[45^\circ \] with the horizontal. It is just able to clear two hurdles of height \[5{\rm{ m}}\] each, separated from each other by a distance d. The value of d is
(a) \[10.4{\rm{ }}\,{\rm{m}}\]
(b) \[18.5{\rm{ m}}\]
(c) \[28.2{\rm{ m}}\]
(d) \[35.6{\rm{ m}}\]

Answer 1

Hint: We will use the expression for vertical distance from the ground which gives us the relationship between horizontal distance from the initial point, the angle of inclination and acceleration due to gravity.

Complete step by step answer:Given:
The speed of the projectile when it leaves the ground is \[U = 20{\rm{ m}}{{\rm{s}}^{ - 1}}\].
The initial angle made by the projectile with the horizontal is \[\theta = 45^\circ \].
The vertical height of the hurdles with the ground is \[h = 5{\rm{ m}}\].

Using the concept of projectile motion, we can write the expression for vertical distance of the hurdles from the ground as below:
\[h = x\tan \theta - \dfrac{{g{x^2}}}{{2{U^2}{{\cos }^2}\theta }}\]……(1)
Here x is the horizontal distance of the first hurdles from the initial point, g is the acceleration due to gravity.

We can write the approximate value of acceleration due to gravity as below:
\[g = 10{\rm{ m}}{{\rm{s}}^{ - 2}}\]

We will substitute \[5{\rm{ m}}\] for h, \[10{\rm{ m}}{{\rm{s}}^{ - 2}}\] for g, \[45^\circ \] for \[\theta \] and \[20{\rm{ m}}{{\rm{s}}^{ - 1}}\] for U in equation (1) and solve it for the value of x.
\begin{align*}
5{\rm{ m}} &= x\tan 45^\circ - \dfrac{{\left( {{\rm{10 m}}{{\rm{s}}^{ - 2}}} \right){x^2}}}{{2{{\left( {20{\rm{ m}}{{\rm{s}}^{ - 1}}} \right)}^2}{{\cos }^2}45^\circ }}\\
\Rightarrow 5{\rm{ m}} &= x\left( 1 \right) - \dfrac{{\left( {10{\rm{ m}}{{\rm{s}}^{ - 2}}} \right){x^2}}}{{2{{\left( {20{\rm{ m}}{{\rm{s}}^{ - 1}}} \right)}^2}{{\left( {\dfrac{1}{{\sqrt 2 }}} \right)}^2}}}\\
\Rightarrow 5{\rm{ m}} &= x - \dfrac{{\left( {10{\rm{ m}}{{\rm{s}}^{ - 2}}} \right){x^2}}}{{{{\left( {20{\rm{ m}}{{\rm{s}}^{ - 1}}} \right)}^2}}}
\end{align*}
We know that x is a distance so its unit is in metre. Writing the above equation in the form of quadratic equation, we get:
\[{x^2} - 40x + 200 = 0\]

We can compare the above equation to get the value of constants with the standard form of a quadratic equation \[{\rm{A}}{x^2} - {\rm{B}}x + {\rm{C}} = 0\], we get:
\[
A = 1\\
B = - 40\\
C = 200
\]

To find the roots of the above equation, we can write the Sridhar-Acharya formula as below:
\[x = \dfrac{{ - {\rm{B}} \pm \sqrt {{{\rm{B}}^2} - 4{\rm{AC}}} }}{{2{\rm{A}}}}\]

We will substitute 1 for A, \[ - 40\] for B and 200 for C in the above expression to find the roots of the quadratic equation.
\begin{align*}
x &= \dfrac{{ - \left( { - 40} \right) \pm \sqrt {{{\left( { - 40} \right)}^2} - 4\left( 1 \right)\left( {200} \right)} }}{{2\left( 1 \right)}}\\
 \Rightarrow x &= \dfrac{{40 \pm \sqrt {800} }}{2}\\
\Rightarrow x &= 20 \pm 10\sqrt 2
\end{align*}

We can write the roots of the above equation as:
\[{x_1} = 20 + 10\sqrt 2 \]
And,
\[{x_2} = 20 - 10\sqrt 2 \]

We can write the expression for the distance between the two hurdles.
\[d = {x_1} - {x_2}\]

We will substitute \[20 + 10\sqrt 2 \] for \[{x_1}\] and \[20 - 10\sqrt 2 \] for \[{x_2}\] in the above expression.
\begin{align*}
d &= \left( {20 + 10\sqrt 2 } \right) - \left( {20 - 10\sqrt 2 } \right)\\
\Rightarrow d &= 20\sqrt 2 \\
\Rightarrow d &= 20\left( {1.414} \right)\\
\Rightarrow d &= 28.28
\end{align*}
We know that the unit of distance is metre.
\[d = 28.28{\rm{ m}}\]

Therefore, the distance between two hurdles is \[28.28{\rm{ m}}\] and option (C) is correct.

Note:We can remember the expression for roots of the equation using Sridhar-Acharya formula. We can also remember the exact value of acceleration due to gravity to find the exact answer which is equal to \[9.81{\rm{ m}}{{\rm{s}}^{ - 2}}\].