Question

Q. A particle is projected with velocity u= 40 m/s at an angle q = 300, with the horizontal from a point on the ground. FindThe change in velocity between the point of projection and point of impact if these points are on the ground ?

Answer 1

When you project a particle with velocity `40 m/s` at angle `30` w.r.t the horizontal,it touches the ground with the same velocity,making the same angle,which is shown below.

Now,if we represent, both vector as `vec A`(where, `|vecA|=40`),then we can say,change in velocity = final velocity with which touching the ground - initial velocity with which it left the ground,

i.e `vec A - vec A` (well the answer is not zero,as the angle between them is not zero)

or,we can write, `vec A +(-vecA)`

Now,angle between `vec A` and `-vec A` is `60` degrees,

So,change in velocity = `sqrt( A^2 +A^2 +2A A cos 60)=sqrt(3)A=40sqrt(3)` (as `A=40 m/s`)