Question

A body is projected at an angle of ${{45}^{o}}$ with KE E. The K.E at the highest point is:
A. zero
B. \[\dfrac{3E}{4}\]
C. \[\dfrac{E}{2}\]
D. \[E\]

Answer 1

Hint: First of all we will find the value of velocity at the highest point of trajectory. Then, substitute that velocity value in the formula for kinetic energy. By further solving the equations we can derive the final answer.
Formula: \[K.E=\dfrac{1}{2}m{{v}^{2}}\]

Complete answer:
When a body is projected at an angle of ${{45}^{o}}$ it will follow a projectile motion.
We know that kinetic energy of a body is given by
$K.E=\dfrac{1}{2}m{{v}^{2}}$
Where m represents mass of the body and v represents the velocity of the body.
As per our knowledge, we know the fact that at the highest point of the trajectory the vertical component of velocity becomes zero and the horizontal component is given by the formula \[u{}_{x}=u\cos \theta \]
Where ‘u’ is the velocity with which the body is projected and angle given in this question is 45\[{}^\circ \].
So let us now substitute $\theta ={{45}^{o}}$ in horizontal component of velocity
So, we get \[{{u}_{x}}=u\cos 45\]
$\Rightarrow {{u}_{x}}=\dfrac{u}{\sqrt{2}}$ as \[\cos 45{}^\circ =\dfrac{1}{\sqrt{2}}\]
In the question it is given that body is projected with kinetic energy E that is
$K.E=E=\dfrac{1}{2}m{{u}^{2}}$
Now, kinetic energy at the highest point will be
\[\begin{align}
  & KE=\dfrac{1}{2}m{{u}_{x}}^{2} \\
 & \Rightarrow \dfrac{1}{2}m{{(\dfrac{u}{\sqrt{2}})}^{2}} \\
 & \Rightarrow \dfrac{1}{2}\times \dfrac{m{{u}^{2}}}{2} \\
 & \Rightarrow \dfrac{E}{2} \\
\end{align}\]
$\therefore K.E=\dfrac{E}{2}$
 So, we get to the conclusion that at the highest point of trajectory the kinetic energy becomes half.

Therefore, Option (C) is correct.

Note:
An object thrown or projected into the air is subject to only the acceleration of gravity. The motion of the object under the acceleration of gravity is known as projectile motion. The object is called a projectile and the path followed by the object is called trajectory. The Range of the projectile is given by the formula $R=\dfrac{{{u}^{2}}\sin 2\theta }{g}$ and the maximum height attained by it given by $H=\dfrac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}$.