Question

A particle of mass m is projected with velocity ‘u’ at an angle θ with horizontal. During the period when the particle descends from the highest point in its trajectory to a point where its velocity vector makes an angle half the angle of projection with the horizontal; the work done by the force of gravity is

(1) -1/2 mu² cos² θ tan² θ/2

(2) +1/2 mu² cos² θ tan² θ/2

(3) 1/2 mu² tan² θ/2

(4) 1/2 mu tan² θ/2 

Answer 1

(2) +1/2 mu² cos² θ tan² θ/2

At the highest point, A, in the trajectory; only horizontal component of velocity u will be present i.e. u cos θ . At P, the instantaneous

velocity v, makes an angle θ/2 with the horizontal.

Horizontal component of v = v cos (θ/2)

As horizontal component of velocity remains constant, we have,

Work done by force of gravity. 

= Change in K.E. from A to B.