A particle is projected at an angle of 45^{ci | vedclass

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(B) The initial kinetic energy is given by $K = \frac{1}{2}mu^2$, where $u$ is the initial velocity.At the highest point of the trajectory, the vertic

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$K/2$

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(B) The initial kinetic energy is given by $K = \frac{1}{2}mu^2$, where $u$ is the initial velocity.At the highest point of the trajectory, the vertical component of velocity becomes zero, and the particle only possesses the horizontal component of velocity, which is $v_x = u \cos 	heta$.The kinetic energy at the highest point $K'$ is given by $K' = \frac{1}{2}m(u \cos 	heta)^2 = \frac{1}{2}mu^2 \cos^2 	heta$.Substituting $	heta = 45^{\circ}$, we get $K' = K \cos^2(45^{\circ}) = K 	imes (1/\sqrt{2})^2 = K/2$.

$A$ particle is projected at an angle of $45^{\circ}$ with the horizontal with an initial kinetic energy $K$. What will be its kinetic energy at the highest point?

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