If α and u are the angle of projection and initial velocity of a projectile respectively, the horizontal range of the projectile, is

ANS:A -

Answer should be U^2sin(2a)/g

As range will be horizontal component of velocity i.e Ux multiply by t(time)____ Ux t= Usin(a)*t
now for finding t let us consider vertical motion of proectile

So, S = ut + 0.5 at^2.
0 = Uy *t - 0.5 * a * t^2 (put Uy as Usin(a)).
Solving this we get t = 2Usin(a)/g.
Hence range R = Ux * t = Ucos(a) * 2 * Usin(a)/g.
Hence R = U^2sin(2a)/g. Horizontal range= (u^2sin(2$))/g
Maximum ht=(u^2sin^2($))/2g
Time to reach maximum ht=(usin($))/g
Time of flight=(2usin($))/g