Question

If R and H are the horizontal range and maximum height attained by a projectile, then its speed of projection is:
A. $\sqrt {2gR + \dfrac{{4{R^2}}}{{gH}}}$
B. $\sqrt {2gH + \dfrac{{g{R^2}}}{{8H}}}$
C. $\sqrt {2gH + \dfrac{{8H}}{{gR}}}$
D. $\sqrt {2gH + \dfrac{{{R^2}}}{H}}$

Answer 1

Start by writing all the relevant formulas related to projection at an angle. Use different trigonometric identities and try to represent or find relation between the two formulas. Rearrange the terms in order to find the desired value in terms of other variables.

Complete answer:
We know the horizontal range and maximum height are given by the formula $R = \dfrac{{{u^2}\sin 2\theta }}{g}$ and $H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$ respectively.
Where u is the initial velocity or velocity of projection, $\theta$ is the angle of projection and g is the acceleration due to gravity.
Now we know the trigonometric identity $\sin 2\theta = 2\sin \theta \cos \theta$, substituting this in range formula, we will get $R = \dfrac{{2{u^2}\sin \theta \cos \theta }}{g}$.
Now dividing height formula by this new range ,we get $\tan \theta = \dfrac{{4H}}{R} \to eqn(1)$
We also know that $\sin 2\theta = \dfrac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}$, substituting value from eqn. (1) into this, we get
$\sin 2\theta = \dfrac{{2\dfrac{{4H}}{R}}}{{1 + \dfrac{{16{H^2}}}{{{R^2}}}}} \\\ \sin 2\theta = \dfrac{{\dfrac{{8H}}{R}}}{{\dfrac{{16{H^2} + {R^2}}}{{{R^2}}}}} \\\ \Rightarrow \sin 2\theta = \dfrac{{8HR}}{{16{H^2} + {R^2}}} \\\$
Substituting this value in range we get
$R = \dfrac{{{u^2}\dfrac{{8HR}}{{16{H^2} + {R^2}}}}}{g} \\\ {u^2} = \dfrac{{g(16{H^2} + {R^2})}}{{8H}} \\\ u = \sqrt {\dfrac{{g(16{H^2} + {R^2})}}{{8H}}} \\\ u = \sqrt {\dfrac{{g16{H^2}}}{{8H}} + \dfrac{{{R^2}}}{{8H}}} \\\ u = \sqrt {2gH + \dfrac{{{R^2}}}{{8H}}} \\\$

So, the correct answer is “Option D”.

Note:
Students must be thorough with all the formulas associated with horizontal and vertical projection such as velocity at any point , time of flight etc. , because many a times its just about placing the values in the formula and re-arrangement. Student’s must also remember the equation of trajectory which is given by the relation $y = x\tan \theta - \dfrac{{g{x^2}}}{{2u{{\cos }^2}\theta }}$, where
Y is the vertical position (m)
x is the horizontal position (m)
u is the initial velocity (combined components, m/s)
g is the acceleration due to gravity (9.80 m/$s^2$)
$\theta$ is the angle of the initial velocity from the horizontal plane (radians or degrees)