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Step-by-Step Solution
Step 1: Identify the Initial Kinetic Energy
The ball is projected with a kinetic energy $E$. By definition, this kinetic energy can be written as
$E = \frac{1}{2} m u^2,$
where $m$ is the mass of the ball and $u$ is the initial speed.
Step 2: Recognize the Velocity Components
At the moment of projection, the velocity has both horizontal and vertical components since it is thrown at an angle of $45^\circ$ to the horizontal:
Horizontal component of velocity: $u \cos 45^\circ = u \,\frac{1}{\sqrt{2}}.$
Vertical component of velocity: $u \sin 45^\circ = u \,\frac{1}{\sqrt{2}}.$
Step 3: Focus on the Velocity at the Highest Point
At the highest point of the projectile's flight, the vertical component of velocity becomes zero (because gravity temporarily brings the upward velocity to zero). Hence, the only velocity component that remains is the horizontal component, which is unaffected by gravity (assuming negligible air resistance).
Thus, at the highest point, the speed of the ball is
$u \cos 45^\circ = \frac{u}{\sqrt{2}}.$
Step 4: Calculate the Kinetic Energy at the Highest Point
The kinetic energy at any point is given by
$\frac{1}{2}\,m\,(\text{speed})^2.$
At the highest point, the speed is $\frac{u}{\sqrt{2}}$. Therefore, the kinetic energy at the top becomes:
$\text{Kinetic Energy at highest point} = \frac{1}{2}\,m\,\left(\frac{u}{\sqrt{2}}\right)^2
= \frac{1}{2}\,m\left(\frac{u^2}{2}\right)
= \frac{1}{2}\,m\,u^2 \times \frac{1}{2}
= \frac{1}{2}E.$
Step 5: Conclude the Final Answer
Thus, the kinetic energy at the highest point is exactly half of the initial kinetic energy, i.e.,
$\frac{E}{2}.$
