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A small block slides with velocity 0.5√gr on the horizontal frictionless surface The block leaves the surface at poibt C . Then angle tita is

Shaikh Abdullah , 8 Years ago
Grade 12
anser 1 Answers
Khimraj
The block leave contact where the normal reaction is zero so mv²/r = mgcosθ and v² = u²+2as so v² = 1/4gr + 2gr(1-cosθ) so 1/4 + 2(1-cosθ) = cosθ so 3cosθ = 9/4 so cosθ = 3/4 and θ = cos-¹(3/4).Hope it clears. If u have doubts then please clearify. And if u like my answer then please approve answer
Last Activity: 8 Years ago
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