Question

If an object is thrown at an angle of ${{60}^{\circ }}$ with horizontal, find the elevation angle of the object at its highest point as seen from the point of projection.
a)${{\tan }^{-1}}\dfrac{\sqrt{3}}{2}$
b)${{\tan }^{-1}}\dfrac{1}{2}$
c)${{\tan }^{-1}}\dfrac{1}{\sqrt{2}}$
d)none of the above

Answer 1

Hint: The angle of elevation is the angle subtended by the line joining the point of projection and the maximum height. The maximum height is the point when the projectile has only horizontal components of velocity. Using trigonometric equations we can accordingly determine the angle of elevation.
Formula used:
$h=\dfrac{{{u}^{2}}\sin \text{ }\!\!\theta\!\!\text{ }}{2g}$
$\dfrac{R}{2}=\dfrac{2{{u}^{2}}\sin \theta cos\theta }{2g}$

Complete answer:

From the above figure it can be observed that the angle of elevation is $\text{ }\!\!\beta\!\!\text{ }$ . Let us say a body is projected at an initial velocity of ‘u’ and at an angle $\text{ }\!\!\theta\!\!\text{ }$ with respect to the horizontal. The body is accelerating downwards with ‘g’ due to gravity. Then the maximum height (h) attained by a body in projectile motion is given by,
$h=\dfrac{{{u}^{2}}{{\sin }^{2}}\text{ }\!\!\theta\!\!\text{ }}{2g}$
Similarly the range (R)of the projectile i.e. the horizontal distance covered during the time of flight is given by,
$R=\dfrac{2{{u}^{2}}\sin \theta cos\theta }{g}$
The maximum height attained by the projectile is exactly at half the range covered. Hence the point on the x-axis at which the height attained is maximum is,
$\dfrac{R}{2}=\dfrac{{{u}^{2}}\sin \theta cos\theta }{g}$
From the figure we can define tangent of beta as,
$\begin{align}
  & \tan \text{ }\!\!\beta\!\!\text{ }=\dfrac{h}{R/2} \\
 & \tan \text{ }\!\!\beta\!\!\text{ }=\dfrac{\dfrac{{{u}^{2}}{{\sin }^{2}}\text{ }\!\!\theta\!\!\text{ }}{2g}}{\dfrac{{{u}^{2}}\sin \theta cos\theta }{g}}=\dfrac{\dfrac{\sin \text{ }\!\!\theta\!\!\text{ }}{2}}{\cos \text{ }\!\!\theta\!\!\text{ }}=\dfrac{\tan \text{ }\!\!\theta\!\!\text{ }}{2} \\
 & \Rightarrow \tan \text{ }\!\!\beta\!\!\text{ }=\dfrac{\tan {{60}^{\circ }}}{2},\because \tan {{60}^{\circ }}=\sqrt{3} \\
 & \Rightarrow \tan \text{ }\!\!\beta\!\!\text{ }=\dfrac{\sqrt{3}}{2} \\
 & \Rightarrow \text{ }\!\!\beta\!\!\text{ }={{\tan }^{-1}}\dfrac{\sqrt{3}}{2} \\
\end{align}$

Therefore the correct answer of the above question is option a.

Note:
In the above question we have ignored the existence of air where the projection of the body happens. Hence the above mathematical quantities hold valid. In reality, the actual range and the maximum height will always be minimum.