A man can throw a stone with initial speed 10\[m\]⁄\[s\]. Find the maximum horizontal distance to which he throw the stone in a room of height \[h\] \[m\] for, (1) \[h = 2m\] and (2) \[h = 4m\].

Hint: Firstly consider the motion is projectile and find the equations of projectile motion. Substitute the values in the equation of motion and find the angle value. After that, use the maximum height formula and calculate the maximum height.

Formula used:
The horizontal range of a projectile motion is given by
\[H\]=\[\dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}\]
Here, \[H\]is maximum height,\[u\]is initial speed and\[g\]is gravity.
\[{H_{\max }}\]= \[\dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}\]
Here \[{H_{\max }}\]is the maximum height.

Complete step by step solution:
Given that, Initial speed of stone = 10 \[m\]⁄\[s\].
(1) At height \[h = 2m\]
\[H\]= \[\dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}\]
Here, \[H\]is maximum height,\[u\]is initial speed and\[g\]is gravity.
 Given, \[h = 2m\]
Substitute given values
 \[2 = \dfrac{{{{\left( {10} \right)}^2} \times {{\sin }^2}\theta }}{{2 \times 10}}\]
$\implies$ \[\sin \theta = \sqrt {\dfrac{2}{3}} \]
So, \[\cos \theta = \sqrt {\dfrac{1}{3}} \]
\[R\]= \[\dfrac{{{u^2}\sin 2\theta }}{g}\] ( ∴\[\sin 2\theta = 2\sin \theta \cos \theta \] )
Here, \[R\]represents range and \[g\]is gravity.
By substituting values
\[R\]= \[\dfrac{{{{\left( {10} \right)}^2} \times 2 \times \left( {\sqrt {\dfrac{2}{3}} } \right)\left( {\sqrt {\dfrac{1}{2}} } \right)}}{{10}}\]
$\implies$ \[R\]= \[4\sqrt 6 \]\[m\]
(2) At height \[h = 4m\]
\[H\]= \[\dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}\]
Here, \[H\]is maximum height,\[u\]is initial speed and\[g\]is gravity.
Given, \[h = 4m\]
Substitute given values
\[4 = \dfrac{{{{\left( {10} \right)}^2}{{\sin }^2}\theta }}{{2g}}\]
$\implies$ \[\sin \theta = \sqrt {\dfrac{4}{5}} \]
  So, \[\cos \theta = \sqrt {\dfrac{1}{5}} \]
$\implies$ \[R\]= \[\dfrac{{{u^2}\sin 2\theta }}{g}\] ( ∴\[\sin 2\theta = 2\sin \theta \cos \theta \] )
Here, \[R\]represents range and \[g\]is gravity.
By substituting values
\[R\]= \[\dfrac{{{{\left( {10} \right)}^2} \times 2 \times \sqrt {\dfrac{4}{5}} \sqrt {\dfrac{1}{5}} }}{{10}}\]
$\implies$ \[R\]= \[8\] \[m\].

Note:
The maximum height and the maximum horizontal range of the object in projectile motion always depends on the angle by which the object is thrown.
The value of the maximum velocity will change if the direction and the angle changes.