- Home
- Standard 11
- Physics
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
Similar Questions
hard