Prove that the time of flight T and the horizontal range R of a projectile are connected by the equation: `gT^2 =2 R tan theta`, where `theta` is the angle of projection.

Find the equation of trajectory, time of flight , maximum height and horizontal range of a projectile when projected at an angle theta with the vertical direction .

The time of flight of a projectile is related to its horizontal range by the equation gT^(2)=2R .The angle of projection is?

Find (a) time of flight , (b) Max.height (c ) Horizontal range of projectile projected with speed (v) making an angle theta with the horizontal direction from ground.

A projectile has the same range R for two angles of projections but same speed. If T_(1) and T_(2) be the times of flight in the two cases, then Here theta is the angle of projection corresponding to T_(1) .

If the time of flight of a bullet over a horizontal range R is T, then the angle of projection with horizontal is -

A projectile is thrown with velocity u making an angle theta with the horizontal. Its time of flight on the horizontal ground is 4 second. The projectile is moving at angle of 45^(@) with the horizontal just one second after the projection. Hence the angle theta is (take g=10m//s^(2) )

Assertion In projectile motion, if time of flight is made two times, then maximum height will become four times. Reason T prop sin theta and H prop sin^(2) theta , where theta is angle of projection.

A projectile has the same range R for angles of projections. If T_(1) and T_(2) be the times of fight in the two cases, then ( using theta as the angle of projection corresponding to T_(1) )

A projectile has a time of flight T and horizontal range R . If the time of flight is 1.5T ,what will be its range now?

A ball is projected with speed u at an angle theta to the horizontal. The range R of the projectile is given by R=(u^(2) sin 2theta)/(g) for which value of theta will the range be maximum for a given speed of projection? (Here g= constant)