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A diwali rocket moves vertically up with...

A diwali rocket moves vertically up with a constant acceleration `a_(1) =20`//` 3 m s^(-2)`. After sometives, its fuel gers exhausted ad then if falls freely with an acceleration `a_(2) =10 m s^(-2)`, If themaximum height attained by the diwalin rocket is (h), using graphicalmerhod, find its speed when the fuel is just exausted. Assume `h=50 m`.

Text Solution

Verified by Experts

Let the rocket acquire a speed `v` at the time when all its fuel gets exhausted. Referring to the `v-t` graph in Fig. the slope of `OP` gives the acceleration `a_(1)`.
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The slope of `OP` gives the acceleration `a_(1) =tan theta_(1) =(v)/(t_(1))`
The slope of `PQ` gives the acceleraion
`(-a_(2))=tan theta_(2) =(-v)/t(2)`
From (i) and (ii), we have
`v=a_(1)t_(1) =a_(2)t_(2)`.
The area under `v-t` graph=Area `Delta OPQ =(1)/(2) OQ.PM =s`
Susbstituting `OQ=t _(1) +t_(2)` and `(t_(1)+t_(2)`
Substituting `t_(12)=(v)/(a_(1))` and `t_(2) (v)/(a_(2))` from (iii) in (iv), we have
`s=(v)/(2)((v)/(a_(1)+(v)/(a_(2))))` Finally substituting `s=+h` (as the area is positive), `a_(1)=(20)/(3) m s^(-2)`
and `a_(2) =10 m s^(-2)`, we have `
`v=sqrt((2a_(1)a_(2)h)/(a_(1)+a_(2)))=sqrt((2xx(20)/(3)xx10xx50)/((20)/(3)+10))`
This sields `v=20 m s^(-1)`.
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