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
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....
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
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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