A gun of muzzle speed `v_(0)` is situated at height h above a horizontal plane. Prove that the angle at which it must be fired so as to achieve the greatest range on the plane is given by-
`theta = (1)/(2)cos^(-1) ((gh)/(v_(0)^(2)+gh))`

If the range of a gun which fires a shell with muzzle speed V is R , then the angle of elevation of the gun is

A particle is projected with a velocity v so that its range on a horizontal plane is twice the greatest height attained. If g is acceleration due to gravity, then its range is

A particle is projeted with a velocity v, so that its range on a horizontal plane is twice the greatest height attained. If g is acceleration due to gravity, then its range is :

If H_(1) and H_(2) be the greatest heights of a projectile in two paths for a given value of range, then the horizontal range of projectile is given by

A cannon ball has the same range R on a horizontal plane for two angles of projection. If h_(1) and h_(2) are the greatest heights in the two paths for which this is possible then:

A projectile is fired with a velocity v at right angle to the slope inclined at an angle theta with the horizontal. The range of the projectile along the inclined plane is

Prove that for a projectile fired from level ground at an angle theta above the horizontal, the ratio of the maximum height H to the range R is given by (H)/(R)=(1)/(4)tan theta

An inclined plane is located at angle alpha = 53^(@) to the horizontal. There is a hole at point B in the inclined plane as shown in the figure. A particle is projected along the plane with speed v_(0) at an angle beta = 37^(@) to the horizontal in such a way so that it gets into the hole. Neglect any type of friction. Find the speed v_(0) (in ms^(-1) ) if h = 1 m and l = 8m.

A horizontal plane 4x-3y+7z=0 is given. Find a line of greatest slope passes through the point (2,1,1) in the plane 2x+y-5z=0