A particle is thrown vertically upward with speed `40 m//s`.
(a) After how much time attains maximum height?
(b) Find (i) the maximum height attained by the particle.
(ii) the velocity at half of the maximum height.
(c ) Find (i) the ratio of distances in the first half and the next half of time.
the ratio of the time in the first half and the second half of the distance (consider only upward journey)

Explanation

Velocity at the highest point, `v=0`
(a) `v=u-g t`
`0=40-10 t_(0)impliest_(0)=4 s`
(b) (i) `v^(2)=u^(2)-2gh`
`0=(40)^(2)-2xx10 H_(max)`
`H_(max)=80 m`
(ii) Velocity at half of the maximum height
`v^(2)=u^(2)-2gh`
`v_(1)^(2)=(40)^(2)-2gH_(max)/2`
`=1600-10xx80=800`
`v_(1)=20sqrt(2) m//s`
(c ) Distance covered by the ball in the first `2 s`,
`h_(1)=ut-1/2g t^(2)=40xx2-1/2xx10xx2^(2)`
`=80-20=60 m`
Distance covered in the next `2 s`,
`h_(2)=H_(max)-h_(1)=80-60=20 m`
`h_(1)/h_(2)=60/20=3/1`
(d) Time taken by the ball to cover the fist `40 m`,
`h=ut_(1)-1/2g t_(1)^(2)`
`40=40t_(1)-5t_(1)^(2)impliest_(1)^(2)-8t_(1)+8=0`
`t_(1)=(8+-sqrt((-8)^(2)-4xx1xx8))/2=(8+-4sqrt(2))/2`
`=4+-2sqrt(2)`
`t_(1)=4+2sqrt(2)` is not possible, because we are interested in upward motion only.
`:. t_(1)=4-2sqrt(2)`
Time in next `40 m`,
`t_(2)=t_(0)-t_(1)=4-(4-2sqrt(2))=2sqrt(2) s`
`t_(1)/t_(2)=(4-2sqrt(2))/(2sqrt(2))=((sqrt(2)-1))/1`