Question

What is the ratio of \[P.E.\] w.r.t. ground and \[K.E.\] at the top most point of the projectile motion:
A. ${\cos ^2}\theta $
B. ${\sin ^2}\theta $
C. ${\tan ^2}\theta $
D. ${\cot ^2}\theta $

Answer 1

Hint:Before we go into the question, let's take a look at projectile motion.The motion of an item hurled or projected into the air, subject only to gravity's acceleration, is known as projectile motion. The object is known as a projectile, and the course it takes is known as a trajectory.

Complete step by step answer:
Height at its highest point \[H = \dfrac{{v_o^2si{n^2}\theta }}{{2g}}\].
\[P.E. = mgH = \dfrac{{mv_o^2si{n^2}\theta }}{2}\]
\[K.E.\] is the only horizontal component at the very top.
Therefore,
\[K.E. = \dfrac{1}{2}m{v^2} \\
\Rightarrow K.E.= \dfrac{1}{2}m{({v_o}cos\theta )^2} \\
\Rightarrow K.E.= \dfrac{1}{2}mv_o^2co{s^2}\theta \]
As a result, the P.E. and K.E. ratios.
\[\dfrac{{\dfrac{{mv_o^2si{n^2}\theta }}{2}}}{{\dfrac{{mv_o^2{{\cos }^2}\theta }}{2}}} = {\tan ^2}\theta \]

Hence, the correct option is C.

Additional Information: The vertical component of a projectile's velocity (at that moment) is zero when it reaches the highest point in its trajectory. As a result, the potential energy at the highest point equals the kinetic energy contribution from the vertical component of beginning velocity.

Note:When investigating projectile motion, it's important to set up a coordinate system. Defining an origin for the \[x\] and \[y\] positions is one aspect of defining the coordinate system. It is frequently more convenient to use the object's initial position as the origin, with \[{x_0} = 0\] and \[{y_0} = 0\] . It's also crucial to distinguish between positive and negative \[x\] and \[y\] directions. The positive vertical direction is commonly defined as upwards, and the positive horizontal direction is usually the direction of motion of the object. When this occurs, the vertical acceleration, \[g\] , becomes negative (since it is directed downwards towards the Earth).