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Asked by Suvarnawoonna | 13 May, 2019, 05:00: PM
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Range of the particle in projectile motion is given by 
R a n g e space equals fraction numerator U squared S i n 2 theta over denominator g end fraction R a n g e space equals fraction numerator 20 squared S i n 2 cross times 45 to the power of 0 over denominator 10 end fraction equals 40 m
As the collision with the walls are perfectly elastic 
Therefore there will be no change in the horizontal velocity and hence the total distance covered by the particle will be 40m
Which is 10 m before collision with first wall, 25 m after collision with first wall and before striking 2nd wall and 5m after collision with second wall. So the particle will land 10 m behind the starting point.
Answered by Utkarsh Lokhande | 14 May, 2019, 12:41: PM
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