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zahid sharief Grade: 11
        

a particle is projected up an inclined plane of inclination beta(b) at an elevation alpha(a) to the inclined plane. find the value of product tan(a).tan(b) if the particle strikes the inclined plane vertically. (ans-0.5)

7 years ago

Answers : (2)

mohit sengar
39 Points
										

as horizontal velocity along the inclined plane is given by


  Vx = ucosa - g sinb (2u sina/g cosb )  as the acc along the inclined plane is given by -gsinb   and  and total time taken is given by 2u sina / gcosb  .


now , as it fall vertically on the inclined plane so Vx = 0


or , u cosa - 2u sina tanb = 0


tan(a ).tan(b) = 1/2


tan(a) . tan(b) = 0.5

7 years ago
Raman Gupta
5 Points
										

take the co-ordinate axis along the inclined plane and Perpendicular to it.


X-direction:->


V=u+at


0=ucos(a)-gsin(b)*t


t=ucosa/gsinb


y direction


-usina=usina-gcosb*t


substitute the value of t, u will get the ans.

7 years ago
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