Question

If R is the range of a projectile on a horizontal plane and h its maximum height, then maximum horizontal range with the same velocity of projection is
A. $2{\rm{h}}$
B. $\dfrac{{{R^2}}}{{8h}}$
C. $2R + \dfrac{{{h^2}}}{{8R}}$
D. $2h + \dfrac{{{R^2}}}{{8h}}$

Answer 1

The above problem can be resolved using the trajectory motion of a projectile. The maximum height attained by the projectile can be defined as the maximum vertical distance that is attained by the body in projectile motion above the plane of horizontal projection. The horizontal range is the total distance that the body in projectile motion travels during its entire time of flight. In this projectile motion, the velocity of the projected body can be divided into two components; the sine component of the velocity is responsible for the maximum height attained by the body under the given conditions. In contrast, the cosine component of the body is the one responsible for the horizontal range.

Complete step by step answer:
According to the question;
Given:
The range of the projectile is R.
The maximum height attained by the body is h.
Now let us consider that the body is projected with a velocity ‘u’ at an angle of $\theta$ with respect to the horizontal. Then,
The range can be written as $R = \dfrac{{{u^2}\sin \;2\theta }}{g}$ …(i)
Height is given as $h = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$ ...(ii)
Here, ‘g’ is the gravitational acceleration of the Earth.
Dividing expression (ii) by expression (i), we get;

 $$ = \dfrac{1}{4}\tan \theta $$  

So, $\tan \theta = \dfrac{{4h}}{R}$
So, the hypotenuse of the triangle of which $\theta$ is an angle is given by;
$x = \sqrt {{R^2} + {{\left( {4h} \right)}^2}} = \sqrt {{R^2} + 16{h^2}}$
Using the above information, we can deduce;
$\sin \theta = \dfrac{{4h}}{x} = \dfrac{{4h}}{{\sqrt {{R^2} + 16{h^2}} }}$ …(iii)
Now, for the maximum range, the value of $\sin \;2\theta$ has to be maximum (i.e., $= 1$ )
So, maximum range, ${R_{\max }} = \dfrac{{{u^2}}}{g}$
Using expression (ii), we have;
$\dfrac{{{u^2}}}{g} = \dfrac{{2h}}{{{{\sin }^2}\theta }}$ …(iv)
Combining expression (iii) and (iv), we get;

{R_{\max }} = \dfrac{{2h}}{{{{\sin }^2}\theta }}\\\ = \dfrac{{2h}}{{{{\left( {4h} \right)}^2}}}\left( {{R^2} + 16{h^2}} \right)\\\ = \dfrac{{2h}}{{16{h^2}}}\left( {{R^2} + 16{h^2}} \right)\\\ = \dfrac{1}{{8h}}\left( {{R^2} + 16{h^2}} \right)\\\ = \dfrac{{{R^2}}}{{8h}} + \dfrac{{16{h^2}}}{{8h}}\\\ = 2h + \dfrac{{{R^2}}}{{8h}} \end{array}$$ Therefore, the maximum horizontal range is $$2h + \dfrac{{{R^2}}}{{8h}}$$ **So, the correct answer is “Option D”.** **Note:** In this question, we need to find the maximum horizontal range with the velocity and the angle of projection for which the maximum height is achieved by the body. First, we need to find out the initial velocity and the sine of the angle of projection, and then we need to put them all together in the expression for maximum horizontal range.