Question
A projectile is projected with kinetic energy $K$. If it has the maximum possible horizontal range, then its kinetic energy at the highest point will be ......... $K$
✓
Answer
since, range is maximum, therefore $\theta=45^{\circ}$
Hence, $v_{x}=v \cos 45^{\circ}=v / \sqrt{2} .$ At the highest
point, the net velocity of the projectile is
$v_{x}=v \cos 45^{\circ}$
$\therefore \quad \mathrm{KE}=\frac{1}{2} \mathrm{mv}_{\mathrm{x}}^{2}=\frac{1}{2} \mathrm{m} \frac{\mathrm{v}^{2}}{2}=\frac{\mathrm{K}}{2}=0.5 \mathrm{K}$
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