A projectile is fired at an angle of 45^o wit | vedclass

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(B) Let the initial velocity be $v$ and the angle of projection be $	heta = 45^o$. The initial kinetic energy is $K = \frac{1}{2}mv^2$.At the maximum

Vedclass · Accepted Answer

$K/2$

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(B) Let the initial velocity be $v$ and the angle of projection be $	heta = 45^o$. The initial kinetic energy is $K = \frac{1}{2}mv^2$.At the maximum height,the vertical component of velocity becomes zero,and the projectile only possesses the horizontal component of velocity,which is $v_x = v \cos 	heta$.The kinetic energy at the maximum height is $K' = \frac{1}{2}m(v_x)^2 = \frac{1}{2}m(v \cos 	heta)^2$.Substituting $K = \frac{1}{2}mv^2$,we get $K' = K \cos^2 	heta$.Given $	heta = 45^o$,we have $\cos 45^o = 1/\sqrt{2}$.Therefore,$K' = K (1/\sqrt{2})^2 = K/2$.

$A$ projectile is fired at an angle of $45^o$ with an initial kinetic energy $K$. What is the kinetic energy of the projectile at its maximum height?

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