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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

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

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    $\begingroup$ 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? $\endgroup$
    – ACuriousMind ♦
    Jul 26, 2014 at 16:09
  • $\begingroup$ @ACuriousMind yes acceleration is same throughout the motion but to calculate the distance between two collinear we must know the velocities between those points $\endgroup$
    – agha rehan abbas
    Jul 26, 2014 at 16:12
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    $\begingroup$ 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. $\endgroup$
    – Tea is life
    Jul 26, 2014 at 16:18
  • $\begingroup$ @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 $\endgroup$
    – agha rehan abbas
    Jul 26, 2014 at 16:21
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    $\begingroup$ initial and final vertical velocities and hence the velocity at the level B are equal. $\endgroup$
    – Tea is life
    Jul 26, 2014 at 16:57

1 Answer 1

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See this page (equation 10, 11 and 12). It will provide the steps needed to show that the equation you have is correct.

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  • $\begingroup$ in the case which i am talking about the body is projected from a tower which is not similar to the case shown in your source $\endgroup$
    – agha rehan abbas
    Jul 26, 2014 at 16:19
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    $\begingroup$ You did not go down to equation 10. Look at it closely, you will see that the levels are different. From the paragraph above "If the projectile is fired at some height h_0 above the ground that it lands on we must find the additional time that it stays in the air." $\endgroup$
    – LDC3
    Jul 26, 2014 at 16:21
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    $\begingroup$ now the point stuck my mind which you are emphasising . thanks for that source and explanation $\endgroup$
    – agha rehan abbas
    Jul 26, 2014 at 16:25

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