A particle is moving in XY-plane. It starts to move from origin O at an angle a with X-axis. It has a constant acceleration in negative Y-direction. After some time it passes through a point B in a direction making angle β with the X-axis. If OB makes θ=45∘ angle with the X-axis, the value of tanα+tanβ is __________.
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answer is 2.
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Detailed Solution
The X-coordinate of point B isx=V0cosαt where, V0 is initial velocity and y=v0tsinα−12a0t2.where, a0 is the acceleration Also, tanθ=yxLet the velocity of the particle is v at point B∴ vcosβ=v0cosα and vsinβ=v0sinα−a0t∴ tanβ=vyvx=v0sinα−a0tv0cosα=tanα−a0tv0cosα .....(i)and tanθ=yx=v0tsinα−12a0t2v0tcosαor 2tanθ=2tanα−a0tv0cosα ......(ii)From Eqs. (i) and (ii), we get2tanθ−tanβ=2tanα−tanαor 2tanθ=tanα+tanβ∴tanα+tanβ=2tan45∘=2