Q1. An aero plane flies horizontally at height h at a constant speed v. An antiaircraft gun fires a shell at the plane when it is vertically above the gun. Show that the minimum muzzle velocity of the shell required to hit the plane is ( v² +2gh)^1/2 at an angle tan^-1(2gh/v)^1/2.

Q2.  A projectile is projected with a velocity u at an angle α to the horizontal, in the vertical plane. If after time t, it is moving in direction making an angle β with the horizontal, prove that:                            gt cosβ = u sin(α-β)?

sorry to do this but i have no nickels

Sn=u+a(n-1\2)

this equ does not pass units homogenity

how can it be?please explain(any one)

@giridharg

this equation Sn=u+a(n-1\2) is dimensionaly correct as Sn denotes the distance travel in nth second .............which automatically means distance per (nth)second .........so Sn has a dimension of speed.....

answer to Q 2

let v be the speed after time t .............as we know there is no force in horizontal direction in projectile motion so ......horizontal component of paticle velocity will always be the same

so u cosa = v cosb

now verticlal component of particle velocity after t time is

vsinb= u sina - gt



use the above two equations to get the answer by eliminating v

.......in above equations a= alpha and b =beta