Two bodies are projected from the same point with same speed in the directions making an angle `alpha_(1)` and `alpha_(2)` with the horizontal and strike at same point in the horizontal plane through a point of projection. If `t_(1)` and `t_(2)` are their time of flights.Then `(t_(1)^(2)-t_(2)^(2))/(t_(1)^(2)+t_(2)^(2))`

Two particles are projected from the same point with the same speed at different angles theta_1 & theta_2 to the horizontal. They have the same range. Their times of flight are t_1 & t_2 respectily

A projectile can have the same range R for two angles of projection. If t_(1) and t_(2) be the times of flight in the two cases:-

Two particles projected form the same point with same speed u at angles of projection alpha and beta strike the horizontal ground at the same point. If h_1 and h_2 are the maximum heights attained by the projectile, R is the range for both and t_1 and t_2 are their times of flights, respectively, then

For a given velocity, a projectile has the same range R for two angles of projection. If t_(1) and t_(2) are the time of flight in the two cases, then t_(1) = t_(2) is equal to

For a given velocity, a projectile has the same range R for two angles of rpojection if t_(1) and t_(2) are the times of flight in the two cases then

Two particles are projected from the same point with the same speed at different angles theta_(1) and theta_(2) to the horizontal. If their respective times of flights are T_(1) and T_(2) and horizontal ranges are same then a) theta_(1)+theta_(2)=90^(@) , b) T_(1) =T_(2)tan theta_(1) c. T_(1) =T_(2)tan theta_(2) , d) T_(1)sin theta_(2)=T_(2)sin theta_(1)

Two particles are projected from ground with same intial velocities at angles 60^(@) and 30^(@) (with horizontal). Let R_(1) and R_(2) be their horizontal ranges, H_(1) and H_(2) their maximum heights and T_(1) and T_(2) are the time of flights.Then