A projectile is fired with velocity `v_0` from a gun adjusted for a maximum range. It passes through two points `P and Q` whose heights above the horizontal are `h` each. Show that the separation of the two points is `(v_0)/(g) sqrt(v_0^2 - 4 gh))`.

A projectile is fired with velocity v_(0) from a gun adjusted for a maximum range.It passes through two points P and Q whose heights above the horizontal are h each.The separation of the two points is

At t=0 a projectile is fired from a point O (taken as origin) on the ground wita speed of 50m//s at an angle of 53^(@) with the horizontal. It just passes two points A and B each at height 75m above horizontal. The horizontal separation between the points A and B

A projectile is fired from the origin with a velocity v_(0) at an angle theta with x-axis. The speed of the projectile at an altitude h is .

A projectile is fired from the origin with a velocity v_(0) at an angle theta with x-axis. The speed of the projectile at an altitude h is .

A projectile is fired with velocity. v_0 at an angle 60^@ with horizontal . At top of its trajectory it explodes into two fragments of equal mass. If one fragment retraces the path then the speed of the other fragment is

A projectile is fired with velocity. v_0 at an angle 60^@ with horizontal . At top of its trajectory it explodes into two fragments of equal mass. If one fragment retraces the path then the speed of the other fragment is

A body thrown vertically upwards from a point with a velocity v_(0) rises to a maximum height and then comes back to the point. Then