The range of a projectile is 100 m. Its kine | vedclass

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(D) The kinetic energy of a projectile is given by $KE = \frac{1}{2}mv^2$,where $v$ is the instantaneous velocity of the particle.Since the horizontal

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$100$

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(D) The kinetic energy of a projectile is given by $KE = \frac{1}{2}mv^2$,where $v$ is the instantaneous velocity of the particle.Since the horizontal component of velocity $(v_x = u \cos 	heta)$ remains constant throughout the flight,the net velocity $v = \sqrt{v_x^2 + v_y^2}$ depends on the vertical component $v_y$.The vertical component $v_y = u \sin 	heta - gt$ is maximum in magnitude at the point of projection $(t=0)$ and at the point of landing ($t=T$,where $T$ is the total time of flight).At the point of projection,the horizontal distance covered is $0 \ m$. At the point of landing,the horizontal distance covered is equal to the range $R = 100 \ m$.Comparing the options,the kinetic energy is maximum at the start and at the end of the trajectory. Since $0 \ m$ is not an option,the correct answer is $100 \ m$.

The range of a projectile is $100 \ m$. Its kinetic energy will be maximum after covering a horizontal distance of: (in $m$)

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