Question

Two stones are projected with the same speed but making different angles with the horizontal. Their ranges are equal. If the angle of projection of one is \[\pi \]/3 and its maximum height is \[{h_1}\]; then the maximum height of the other will be
A. 3${h_1}$
B. 2${h_1}$
C. ${h_1}$/2
D. ${h_1}$/3

Answer 1

Hint: If the range of two projectiles are equal then the angle of projection of these projectile ($\theta and\alpha $) are complementary $\theta + \alpha = 90^\circ $ For a particle in projectile motion the maximum height is given as H$ = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$, where u is the initial velocity, g is acceleration due to gravity and $\theta $ is the angle of projection.

Complete step by step answer:
Given that the speed of the two stones projected is equal and let it be u.
The angle of projection of the first stone is $\theta = \pi /3$.
Since the range of the two projectiles are equal therefore the angle of projection of these stones should be complementary, $\theta + \alpha = \pi /2$. Therefore the angle of projection of the second projectile is $
  \alpha = \dfrac{\pi }{2} - \theta \\
  \Rightarrow \alpha = \dfrac{\pi }{2} - \dfrac{\pi }{3} \\
    \Rightarrow \alpha = \dfrac{\pi }{6} \\
 $
Maximum height of a projectile is given as H$ = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
Given the maximum height of the first projectile is ${h_1}$ $ = \dfrac{{{u^2}{{\sin }^2}\dfrac{\pi }{3}}}{{2g}} = $$\dfrac{{{u^2}{{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2}}}{{2g}} = \dfrac{{3{u^2}}}{{8g}}$

Therefore maximum height of the second projectile is ${H'}$ $= \dfrac{{{u^2}{{\sin }^2}\alpha }}{{2g}} = \dfrac{{{u^2}{{\sin }^2}\left( {\dfrac{\pi }{6}} \right)}}{{2g}}$
${H'}$ $= \dfrac{{{u^2}{{\left( {\dfrac{1}{2}} \right)}^2}}}{{2g}} = \dfrac{{{u^2}}}{{8g}} = \dfrac{{{h_1}}}{3}$
Therefore the maximum height of the second projectile is $\dfrac{{h_1}}{3}$.

Note:We say that two projectiles have equal ranges when the angle of projection of the two projectiles are complementary, that is the sum of the angle of projection is equal to 90$^\circ $. Students should learn simple formulae related to projectile motion to solve such types of problems.