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(D) The average power delivered by a force is defined as the work done by the force divided by the time taken.$P_{av} = \frac{W_g}{t}$At the highest p

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$\frac{mgu \sin 	heta}{2}$

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(D) The average power delivered by a force is defined as the work done by the force divided by the time taken.$P_{av} = \frac{W_g}{t}$At the highest point, the vertical displacement is the maximum height $H_{\max} = \frac{u^2 \sin^2 	heta}{2g}$.The work done by gravity $W_g$ as the object moves from the ground to the highest point is $-mgH_{\max}$.The time taken to reach the highest point is $t = \frac{u \sin 	heta}{g}$.Thus, the magnitude of average power is:$P_{av} = \frac{mg H_{\max}}{t} = \frac{mg \left( \frac{u^2 \sin^2 	heta}{2g} ight)}{\left( \frac{u \sin 	heta}{g} ight)}$$P_{av} = \frac{mg u \sin 	heta}{2}$

An object of mass $m$ is projected with an initial velocity $u$ at an angle $\theta$ with the horizontal. What is the average power delivered by gravity in reaching the highest point?

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