the distance travelled by a body during last second of its total flight is 'd'when the body is projected vertically up with certain velocity . If the velocity of projection is doubled, the distance travelled by the body during last second of its total flight is

Asked by sowmya subathra | 17th Apr, 2013, 08:28: PM

Expert Answer:

 Let the total time of flight in first case in t.....

 After reaching the highest point particle will have velocity equal to zero.....and will take time t/2 to come to the ground....

 now t=2u/g   

 so time to come down from peak = u/g

 

Distance travelled in t=u/g will be  u2/2g...

 and in (t-1 )  it will be  g(t-1)2/2  = (u-g)2/2g..

 so d = u2/2g - (u-g)2/2g

 d = (2u-g)/2 --------------(1)

and if projection velocity is doubled.....then we get D= (4u-g)/2

 

from (1)

2d = 2u-g

 

4u = 2(2d+g)

 

4u - g = 4d + g 

surely  D = 2d + g/2

 

Answered by  | 23rd Apr, 2013, 11:46: AM

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