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A body is thrown with a velocity v0 at a...

A body is thrown with a velocity `v_0` at an angle `theta_0` with horizontal. Find the (a) instantaneous power delivered by gravity after a time t measured from the instant of projection and (b) average power delivered by gravity during the time t.

Text Solution

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a. Power delivered by gravity after a time t when the particle has velocity v is,
`P_g=mvecg*vecv`
Substituting `vecg=-ghatj`
and `vec=v_0costheta_0hati+(v_0sintheta_0-g t)hatj` in Eq. (i), we have
`P_g=-mg(v_0sintheta_0-g t)`

b. The average power is
`P_(av)=(underset0oversettintPdt)/(t)`
Substituting `P=P_g=-mg(v_0sintheta_0-g t)`,
we have `P_(av)=-(mg)/(t)underset0oversettint(v_0sintheta_0-g t)dt`
This gives `P_(av)=-mg(v_0sintheta_0-(g t)/(2))`
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