Question

A body of mass m is projected from ground with speed u at an angle theta with horizontal

Answer 1

question is incomplete , A complete question is ----->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 height from ground is ?

solution :- we have to find power delivered by gravity so, use just vertical component.

A body of mass m is projected with speed u at an angle $\theta$ with horizontal.
so, maximum height , $H=\frac{u^2sin^2\theta}{2g}$

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

vertical component of initial velocity , $u_y=usin\theta$

now, use formula, $v_y^2=u_y^2+2a_yy$
so, $v_y^2=(usin\theta)^2-2g\frac{u^2sin^2\theta}{4g}$

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

$v_y=\frac{usin\theta}{\sqrt{2}}$

so, power , $P=F_y.v_y$
here, Fy is the wight of body e.g., Fy = mg

$P=\frac{mgusin\theta}{\sqrt{2}}$

Answer 2

Answer:

Refer to the attachment.