Question

25. A projectile of mass m is thrown with speed u at
an angle 'o' from the horizontal. The moment of the
gravitational force on the projectile about point of
projection, t time after the projection, is
(1) mgut
(2) mgusin0t
(3) mgutcos0
(4) mgutan0t​

Answer 1

Answer:

See horizontal distance travelled by the particle from the point of projection in time t =u*cos(theta)*t

So, moment on the particle by gravitational force about the point of projection = r*F*sin(theta)

r= u*cos(theta)*t

F= mg

Angle is 90° between r and F , so sin(90°) =1

So, , moment on the particle by gravitational force about the point of projection = r*F*sin(theta)= mgutcos(theta)

Here I have not taken vertical distance travelled by the particle since moment of force will be zero in that case as force of gravity too acts vertically. So angle will be 180 and sin(180) =0

Explanation: