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

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 a velocity u in such a way that its horizontal range is three times the maximum height aatained . Find the value of horizontal range .

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 .

two particles are projected upwards with the same initial velocity v_(0) in two different angles of projection such that their horizontal ranges are the same. The ratio of the heights of their horizontal ranges are the same. The ratio of the heights of their highest point will be

Two projectiles thrown from the same point at angles 60^@ and 30^@ with the horizontal attain the same height. The ratio of their initial velocities 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 distance (in metre) of the particle from origin at t=2s

A projectile is projected and it takes 9 s to reach in the horizontal plane through the point of projection. In its path it passes a point P after 4 s . The height of P above the horizontal plane is