in the question number 4, if the horizontal displacement of the particle as seen by an observer on the ground is zero, the speed of the box with respect to the ground at the instant when particle was projected is

in the question number 4, if the horizontal displacement of the partic

1.	in the question number 4, if the horizontal displacement of the particle as seen by an observer on the ground is zero, the speed of the box with respect to the ground at the instant when particle was projected is
Answer» ` U cos alpha ` `(u sin alpha)/(cos theta)` `(u cos (alpha + theta))/(cos theta)` `(u sin theta sin alpha)/(cos theta)` Solution :Horizontal displacement of particle with respect to ground is zero. This implies the there is no horizontal component of the absolute VELOCITY of the particl. Let v be the velocity of the block down the plane. Velocity of particle `=ucos(alpha+theta)hati+USIN(alpha+theta)hatj` Velocity of block `=-vcosthetahati-vsinthetahatj` `:.` Velocity of particle with respect to ground `=[ucos(alpha+theta)-vcostheta]hati+[usin(alpha+theta)-vsintheta]hatj` Now, as said EARLIER that horizontal compoent of absolute velocity should be zero. Therefore, `ucos(alpha+theta)-vcostheta=0` or `v=(ucos(alpha+theta))/(costheta)` (down the plane)