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

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,

\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).