A particle is projected with a velocity u in horizontal direction as shown in the figure. Find u(approx.) so that the particle collides orthogonally with the inclined plane of the fixed wedge.
10m/s
20m/s
None of these
10√2m/s
A
20m/s
B
10√2m/s
C
10m/s
D
None of these
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Solution
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Considering plane along Wedge axis, when particle collides with the plane, its component of velocity parallel to inclined plane should be zero. ⇒ucosθ−gsinθ×t=0 {Equation along the plane} t=ugtanθ u=gt×tanθ=gt√3 Now, Consider the plane along original x-y axis, Thus, Horizontal displacement before it hits the plane =ut vertical displacement during the time t =12gt2 Now , tan30∘=h−12gt2ut 1√3=h−12gt2ut Hence , 13=(hgt2−12)
t=√6h5g Hence , t=2.5938 s So, u=gttan30∘=9.8×2.5938√3=14.69≃10√2 m/s
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