A projectile aimed at a mark, which is in the horizontal plane through the point of projection, falls a cm short of it when the elevation is `alpha` and goes b cm far when the elevation is `beta`. Show that, if the speed of projection is same in all the cases the proper elevation is
`1/2 sin^(-1) [(bsin2alpha+asin2beta)/(a+b)]` .

A projectile is aimed at a mark on a horizontal plane through the point of projection and falls 6 m short when its elevation is 30° but overshoots the mark by 9 m when its elevation is 45^@ . The angle of elevation of projectile to hit the target on the horizontal plane is :

A projectile is aimed at a mark on a horizontal plane through the point of projection and falls 6 m short when its elevation is 30° but overshoots the mark by 9 m when its elevation is 45^@ . The angle of elevation of projectile to hit the target on the horizontal plane is :

A projectile is aimed at a mark on the horizontal plane through the point of projection and falls 12 m short when the angle of projection is 15^(@) , white it overshoots the mark by 24 m when the angle of projection is 45^(@) . Find the angle of projection to hit the mark.

A projectile is required to hit a target whose co-ordinates relative to horizontal and vertical axes through the point of projection are (alpha, beta) If the projection speed is sqrt(2g alpha) . Then the condition when it is impossible to hit the target is

A perefectly elastic particle is projected with a velocity v in a vertical plane through the line of greatest slope of an inclined plane of elevation alpha . If after striking the plane , the particles rebounds vertically show that it will return to the point of projection at the end of time equal to (6v)/(gsqrt(1+8sin^(2) alpha))

The length of the shadow of a person s cm tall when the angle of a elevation of the sun is alpha is p cm. It is q cm when the angle of elevation of the sum is beta . Which one of the following is correct when beta = 3 alpha ?

A parrticle projected with velocity u strikes at right angles a plane through the point of projection inclined at angle beta to the horizon. Show that the height of the point struck above the horizontal plane through the point of projection is (2u^2)/g . (sin^2 beta)/(1+3 sin^2 beta) and that the time of flight up to that instant is t=(2u)/(g sqrt(1+3 sin^2 beta)) .