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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 sometimes, its fuel gets exhausted and then it falls freely with an acceleration `a_(2) =10 m s^(-2)`, If the maximum height attained by the diwali rocket is (h), using graphical method, find its speed when the fuel is just exhausted. Assume `h=50 m`.

Text Solution

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During its ascent with the fuel its motion is upward accelerating. Hence, the slope of `v - t` graph is positive during time `t_(1)`. After its fuel gets exhausted, it accelerates down with `a = g = 9.8 m//s^(2)` and then the slope of `v - t` graph is negative.
The total displacement `= s=` ara under `v - t` graph
`= (v)/(2) (t_(1) + t_(2))`
Since the slope of `v - t` grpah = acceleration, ltbr. `a_(1) = (v)/(t_(1))` .....(ii)
and `a_(2) = (v)/(t_(2))`....(iii)
Using eqs. (i), (ii) and (iii), we have
`v = sqrt((2a_(1)a_(2)h)/(a_(1) + a_(2))) = sqrt((2 xx (20)/(3) xx 10 xx 50)/((20)/(3) + 10))`
or `v = 20 m//s`
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