Question

A particle of mass M is projected with speed u at angle theta with horizontal from ground the work done by gravity on it during its upward motion is

Answer 1

workdone = - change in potential.

initial potential energy,$P.E_i$ = Mgh = M(-g) × 0 = 0

final potential energy = MgH
where H is maximum height of projectile.
e.g., $H=\frac{u^2sin^2\theta}{2g}$
so, final potential energy , $P.E_f=-\frac{Mgu^2sin^2\theta}{2g}$

so, workdone = - change in potential
= - $P.E_f-P.E_i$

= $-\frac{Mgu^2sin^2\theta}{2g}-0$

=$\frac{Mgu^2sin^2\theta}{2g}$

Answer 2

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