Physics, asked by Kumarvikram967, 1 year ago

A diwali rocket moves vertically up with a constant acceleration of 8m/s2. After some time, its fuel gets exhausted and then falls freely under gravity. If maximum height attained by the rocket is 180 m then calculate the speed when the fuel is just exhausted. (g= 10m/s2)​

Answers

Answered by Anonymous
22

Solution:

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Let rocket acquires a speed v at the time when fuel gets exhausted. (Then referring the v-t graph)

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\sf{a_{1} = tan \theta_{1} = \dfrac{v}{t_{1}}}

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Similarly, \sf{tan \theta_{2} = \dfrac{-v}{t_{1}} = a_{2}}

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From above equations, we have,

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v = \sf{a_{1} t_{1} = a_{2} t_{2}}

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Area under (v - t) graph = Area of ∆OPQ \sf{\implies \dfrac{1}{2} OQ \times PN}

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S = \sf{\dfrac{v}{2} (t_{1} + t_{2})}

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\longrightarrow\sf S = \dfrac{v}{2} \bigg(\dfrac{v}{a_{1}} + \dfrac{v}{a_{2}} \bigg)

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\longrightarrow\sf i. e. \: S = \dfrac{v^2}{\: 2} \bigg(\dfrac{a_{1} + a_{2}}{a_{1} a_{2}} \bigg)

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\longrightarrow\sf v = \sqrt{\dfrac{2a_{1} a_{2}}{a_{1} + a_{2}}}

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\longrightarrow \sqrt{\dfrac{2 \times 8 \times 10 \times 180}{18}}

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\longrightarrow\sf{\blue{v = 40 m/s}}

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Hence, speed of rocket when fuel is just exhausted will be 40 m/s.

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