A projectile is fired at an angle of 30° to the horizontal with the speed v. If another projectile is projected with the same speed, then at what angle with the horizontal it must be projected so as to have the same range ?

Particle A is projected with speed u at an angle theta with the horizontal . Another particle B is also projected with same speed u at an angle 90-theta with the horizontal . Both the projectiles are fired from the ground and reaches the ground after completing the projectile . Time of flight for A is t_(A) and that for B it is t_(B) . Range of the projectile A is R_(A) and that for B is R_(B) . Maximum height reached by the projectile A is h_(A) and that for B is h_(B)

(A projectile projected at an angle 30^(@) from the horizontal has a range R . If the angle of projection at the same initial velocity be 60^(@) , then the range will be

A projectile is projected with a speed u at an angle theta with the horizontal. What is the speed when its direction of motion makes an angle theta//2 with the horizontal

Velocity of a projectile is 10 ms^(-1) . At what angle to the horizontal should it be projected so that it covers maximum horizontal distance ?

A projectile thrown with an initial speed u and the angle of projection 15^@ to the horizontal has a range R . If the same projectile is thrown at an angle of 45^@ to the horizontal with speed 2 u , what will be its range ?

When a particle is projected at an angle to the horizontal, it has range R and time of flight t_(1) . If the same projectile is projected with same speed at another angle to have the saem range, time of flight is t_(2) . Show that: t_(1)t_(2)=(2R//g)

A projectile is fired at an angle of 45^(@) with the horizontal. Elevation angle of the projection at its highest point as seen from the point of projection is

A particle is projected with speed 'u' at an angle 60^@ with horizontal, then find the speed when its angle with horizontal changes to 30^@ .

A projectile is projected with speed u at an angle theta with the horizontal . The average velocity of the projectile between the instants it crosses the same level is

Statement-1 : A man projects a stone with speed u at some angle. He again projects a stone with same speed such that time of flight now is different. The horizontal ranges in both the cases may be same. (Neglect air friction) Statement-2 : The horizontal range is same for two projectiles projected with same speed if one is projected at an angle q with the horizontal and other is projected at an angle (90^(@) theta) with the horizontal. (Neglect air friction)