Gujarati
Hindi
3-2.Motion in Plane
medium

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 ......... $K$

A

$0.25$

B

$0.5$

C

$0.75$

D

$1.0$

Solution

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

$v_{x}=v \cos 45^{\circ}$

$\therefore \quad \mathrm{KE}=\frac{1}{2} \mathrm{mv}_{\mathrm{x}}^{2}=\frac{1}{2} \mathrm{m} \frac{\mathrm{v}^{2}}{2}=\frac{\mathrm{K}}{2}=0.5 \mathrm{K}$

Standard 11
Physics

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