# How to derive this formula? [closed]

as far as i know the range of an obliquely projected body which is projected with a velocity of u with some angle $\theta$ with the horizontal under the action of gravity is R=$u^2sin2\theta$/g by which we can calculate the range of projected body when it comes to the same level of the point of projection

but how to calculate the remaining distance which it covered below the point of projection

i got this formula on the web but i cant prove this

Can any one help me by suggesting me what concept I am missing to calculate the range?

• What have you tried? You know the acceleration (from gravity) at every point, and you have the starting velocity and point, so you can simply solve the equation of motion. Where's your problem with that? Jul 26 '14 at 16:09
• @ACuriousMind yes acceleration is same throughout the motion but to calculate the distance between two collinear we must know the velocities between those points Jul 26 '14 at 16:12
• find the range at the same level as B and then find the range from the last point at that level to the final point below. You know the horizontal components of velocity. Add the two ranges. Jul 26 '14 at 16:18
• @learner +1 for this comment but how to find out the velocity along y-axis at the point b where the particle reaches same level of point of projection Jul 26 '14 at 16:21
• initial and final vertical velocities and hence the velocity at the level B are equal. Jul 26 '14 at 16:57