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# A small block of mass m (= 1 kg) is pulled on a frictionless horizontal surface by means of a light rope at a rateof u(=10 m/s) with a pulley situated at a height h(= 9 m) above the level of block. The block was initially placedat a very large distance from the pulley. The angle ? (in degree) of rope with horizontal, when the block leaves thesurface is

Arun Kumar IIT Delhi
7 years ago
Hi,

if it lifts then there will be a vertical component of vu helping it.
$mv_ydv_y/dy>mg \\=>10sin(\theta)*10cos(\theta)d\theta /dt>10 \\y=zsin(\theta) \\=>dy/dt=dz/dt sin(\theta)+zcos(\theta)zd\theta/dt=0 \\=>d\theta/dt=-tan(\theta)*10/z=-tan(\theta)*10*sin(\theta)/y\\ taking\,absolute \,values\\ 10sin(\theta)*10cos(\theta)*tan(\theta)*10*sin(\theta)/y>10 \\=>sin^3(\theta)>9/100$

Thanks & Regards
Arun Kumar
IIT Delhi