Question

A body of mass m is projected from ground with speed u at an angle $\theta$ with horizontal the power delivered by gravity to it at half of maximum heigh from ground is

Answer 1

A body of man 'm' projected with speed U at angle $\theta$

with horizontal

So maximum height $H=\frac{u^2{\sin }^2\theta }{2g}$

half of maximum height $\frac{H}{2}=\frac{u^2{\sin }^2\theta }{4g}$

vertical component of initial velocity $u_y=u\sin \theta$

$v_y^2=u_y^2+2a_{y}y$

$\Rightarrow v_y^2=(u\sin \theta {)}^2-29(\frac{u^2{\sin }^2\theta }{49})$

$v_y^2=\frac{u^2{\sin }^2\theta }{2}=v_y=\frac{u\sin \theta }{\sqrt{2}}$

Power = $F_y{V}_y$

$F_y=mg$

$=\frac{mgu\sin \theta }{\sqrt{2}}$