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A stone is thrwon vertically upwards. On...

A stone is thrwon vertically upwards. On its way is passes point A with a speed v and passes point B, 30 m higher than A with speed `(v)/(2)`. Find (i) The speed v (ii) The maximum height reached by the stone above point B.

Text Solution

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Let us choose the positive direction along vertically upward.
(i) Between the point A and B, using equation `v^(2)=u^(2)+2as`, we get
`(+(v)/(2))^(2)=(+v)^(2)+2(-g)(+30)`
`(v^(2))/(4)=v^(2)-60g`
along the `60g=v^(2)-(v^(2))/(4)=(3)/(4)v^(2)`

`(3)/(4)v^(2)=60g`
`rArr" "v^(2)=60xx(4)/(3)xxg`
`v^(2)=80xx9.8=8xx98`
`v^(2)=4xx2xx2xx49`
`v=sqrt(4xx2xx2xx49)=2xx2xx7=28m//s`
(ii) The speed at `B=(v)/(2)=(28)/(2)=14m//s`
Again using the equation, `v^(2)=u^(2)+2as` between B and C
`0=(+(v)/(2))^(2)+2(-g)(+s)`
`2gs=(14)^(2)=14xx14`
`s=(14xx14)/(2xx9.8)=(14xx14xx10)/(2xx2xx49)=(40)/(4)=10m`
The maximum height attained above B is 10 m.
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