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Question

A particle is projected vertically up from the top of a tower with velocity 10m/s. It reaches the ground in 5s . Find-a) Height of towerb)Striking velocity of particle at groundc)Distance traversed by particle d)Average speed & average velocity of particle

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Solution

When we project the ball from the tower then it'sinitial velocity =u=10m/sfinal velocity=v=0m/s (when it reaches maximum height)acceleration due to gravity=a=g=−9.8m/s²(here - sign because object is moving opposite direction of acceleration)distance =s=?v²−u²=2ass=v²−u²/2as=(0)²−(10)²/2(9.8)s=0−100/−19.6s=−100/−19.6s=5.102m____________________________________________________________When we project the ball from tower then it reaches maximum height then it will be in rest at that moment.Then it starts to fall down intial velocity = u = 0m/s acceleration due to gravity=a=g=9.8m/s²time =t=5sdistance =s2=?s2=ut+at²/2s2=0(5)+9.8(5)²s2=0+9.8(25)s2=245m_________________________________________so total distance traveled by an objectS3=s1+s2S3=245+5.102S3=250.102mso total distance covered by an object is 250.102 meter___________________________________________If you want to find the height of the tower then ,h=S2−s1h=245−5.012h=239.898 meter

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