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A particle is projected making angle $45^\circ$ with horizontal having kinetic energy K. The kinetic energy at highest point will be : -
A
$\dfrac{K}{\sqrt{2}}$
B
$\dfrac{K}{2}$
C
2K
D
K
Detailed Solution
At the highest point, the vertical component of velocity is zero and only the horizontal component $u\cos\theta$ remains.
$K' = \dfrac{1}{2}m(u\cos\theta)^2 = K\cos^2\theta$
$K' = K\cos^2 45^\circ = K\left(\dfrac{1}{\sqrt{2}}\right)^2 = \dfrac{K}{2}$
$K' = \dfrac{1}{2}m(u\cos\theta)^2 = K\cos^2\theta$
$K' = K\cos^2 45^\circ = K\left(\dfrac{1}{\sqrt{2}}\right)^2 = \dfrac{K}{2}$