# If a projectile is projected with kinetic energy.having maximum possible horizontal range.then what will be it's kinetic energy at highest point?

Aug 6, 2018

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Aug 7, 2018

kinetic energy at highest point is $\frac{1}{2} \cdot m \cdot {\left({v}_{i} \cdot \cos \theta\right)}^{2}$

#### Explanation:

At highest point of any projectile, the vertical component of its velocity has decayed to zero. Therefore, its only velocity at that point is its horizontal component (and the kinetic energy will be at the minimum value of its flight). To be projected for highest range, the elevation, $\theta$, of the launch must have been ${45}^{\circ}$.

Let the initial velocity be ${v}_{i}$. Nothing is said about air resistance, so we have to assume there is none. Therefore the horizontal component, ${v}_{\text{hi}}$, does not change for the entire flight.

${v}_{\text{hi}} = {v}_{i} \cdot \cos$

Therefore its kinetic energy at the highest point will be

$K {E}_{\text{min}} = \frac{1}{2} \cdot m \cdot {\left({v}_{i} \cdot \cos \theta\right)}^{2}$

I hope this helps,
Steve