Show that for a given velocity of projection, there are two angles of projection which give the same range of projectile.

Define the terms : (i) projectile (ii) Velocity of projection (iii) Angle of projection (iv) Trajectory of projectile.

The maximum height attained by projectile is increased by 10%, by changing the angle of projection, without changing the speed of projection. The percentage increase in the time of flight will be

The time (T) required by the projectile to retum to original plane of projection is given by

A projectile is fired from the surface of the earth with a velocity of 7 ms^-1 and angle theta with the horizontal. Another projectile fired from another planet with a velocity of 2.5 ms^-1 at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is (in ms^-1 (given =9,8 ms^-2 )

A body is projected at an angle of projection of 35^@ with horizontal. To get the same range with the same velocity of projection, the body should also be projected at an angle of

For a projectile motion, the horizontal range of projectile is same for any two angles which are

If the equation of a projectile is y=sqrt3x - gx^2/2 , then the angle of projection is

The angle of projection for a projectile thrown parallel to horizontal is

The relative velocity of two objects moving with same speed and in the same direction is

The range of a projectile, when launched at an angle of 15^@ with the horizontalis 1.5 km. What is the range of the projectile, when launched at an angle of 45^@ with the horizontal?