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Q .1- A projectile projected at some angle with vel. 50m/s crosses a 20m wall after 4s from the time of projection. The angle of projection of particle will be ? ( pls give full sol.)                                                                                   2. a swimmer can swim in still water at 1.5m/s , the river is flowing with 2m/s from left to right. The minimum time taken by swimmer to cross the river and drift will be? ( pls full sol. )

Sumit Majumdar IIT Delhi
6 years ago
Solution:
1. Let the angle of projection be $\Theta$. Then using the expression of vertical height covered, we get:
$20=\frac{50^{^{2}}sin^{2}\Theta }{2g}$
On solving this we get:
$sin\Theta =\frac{2}{5}$
So, $\Theta =sin^{-1}\left ( {\frac{2}{5}} \right)$
1. The distance between the left to right end of the river isnt given.
Thanks & Regards
Sumit Majumdar,