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Q.

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+tan⁡β is __________.

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answer is 2.

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Detailed Solution

The X-coordinate of point B isx=V0cos⁡αt  where, V0 is initial velocity and y=v0tsin⁡α−12a0t2.where, a0 is the acceleration Also, tan⁡θ=yxLet the velocity of the particle is v at point B∴     vcos⁡β=v0cos⁡α and     vsin⁡β=v0sin⁡α−a0t∴     tan⁡β=vyvx=v0sin⁡α−a0tv0cos⁡α=tan⁡α−a0tv0cos⁡α   .....(i)and tan⁡θ=yx=v0tsin⁡α−12a0t2v0tcos⁡αor 2tan⁡θ=2tan⁡α−a0tv0cos⁡α    ......(ii)From Eqs. (i) and (ii), we get2tan⁡θ−tan⁡β=2tan⁡α−tan⁡αor   2tan⁡θ=tan⁡α+tan⁡β∴tan⁡α+tan⁡β=2tan⁡45∘=2
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