Step 1: Understanding the Concept:
In ideal projectile motion, the horizontal component of velocity remains constant throughout the flight, while the vertical component changes due to gravity.
At the highest point, the vertical component becomes momentarily zero.
Step 2: Key Formula or Approach:
Initial Kinetic Energy $E = \frac{1}{2}mv^2$.
Initial horizontal velocity $v_x = v\cos\theta$.
Velocity at the highest point $v_{\text{top}} = v_x = v\cos\theta$.
Kinetic energy at highest point $E' = \frac{1}{2}m(v_{\text{top}})^2$.
Step 3: Detailed Explanation:
Let mass be $m$ and initial velocity be $v$.
The initial kinetic energy is:
\[ E = \frac{1}{2}mv^2 \]
At the highest point, the projectile has no vertical velocity. Its only velocity is the horizontal component, which has not changed.
\[ v' = v\cos\theta \]
The kinetic energy $E'$ at this highest point is:
\[ E' = \frac{1}{2}m(v')^2 \]
Substitute the expression for $v'$:
\[ E' = \frac{1}{2}m(v\cos\theta)^2 \]
\[ E' = \frac{1}{2}mv^2 \cos^2\theta \]
Recognizing that $\frac{1}{2}mv^2$ is the initial kinetic energy $E$, we substitute:
\[ E' = E \cos^2\theta \]
Step 4: Final Answer:
The kinetic energy is $E \cos^2 \theta$.