Concept of projectile motion

Question

Two second after projection, a projectile is moving at !!30^0!! above the horizontal; after one more second it is moving horizontally. Then:

a) Angle of projection is !!30^0!!

b) Velocity of projection is !!20sqrt(3) m//s!!

c) Velocity at any time will be !!40 m//s!!

d) Maximum horizontal range attained !!= 60sqrt3 m!!

Solution

Initial velocity !!= u!! at angle !!theta!!

Time of flight:

!!u_x = u costheta, and u_y= u sintheta!!

!!(2usintheta)/g = 2(2+1)!!

!!=> u_y = 30 m//s!!

After !!2!! sec, velocity !!=v!! at angle !!30^0!!

!!vcos30^0 = u_x!!

!!vsin30^0= u_y-g(2)=> v = 20 m//s!! and !!u_x = 10sqrt3 m//s!!

!!tantheta= u_y/u_x= 30/(10sqrt3)= sqrt3!!

!!=> theta = tan^-1(sqrt3)= 60^0!!

!!u=sqrt(u_x^2+u_y^2)= 20sqrt3!!

!!R = (u^2sin120^0)/g = 1200/g(sqrt3/2)= 60sqrt3 m!!

Therefore, option (B) and (D) are correct.