A rocket is launched at an angle $53^{0}$ to the horizontal with an initial speed of $100 m s^{-} 1$ . It moves along its initial line of motion with an acceleration of $30 m s^{-} 2$ for 3 seconds. At this time its engine falls & the rocket proceeds like a free body. Find:
(i) the maximum altitude reached by the rocket,
(ii) total time of flight,
(iii) the horizontal range.

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Solution

$S p e e d o f t h e r o c k e t a f t e r 3 s e c o n d s = 100 + 30 \cdot 3 = 190 m / s d i s p l a c e m e n t = 100 \cdot 3 + \frac{1}{2} \cdot 30 \cdot 3^{2} = 435 m v e r t i c a l d i s p l a c e m e n t = 435 s i n 53^{0} = 348 m h o r i z o n t a l d i s p l a c e m e n t = 435 c o s 53^{0} = 261 m a_{x} = 0^i u_{x} = 114^i a_{y} = - g^j u_{y} = 152^j H = \frac{u_{y}^{2}}{2 g} = 649.8 m m a x A l t i t u d e = 649.8 + 348 = 997.8 m t i m e o f f l i g h t u n d e r g r a v i t y i s g i v e b y - 348 = 152 T - \frac{1}{2} g T^{2} \Rightarrow T = 32.5 s t o t a l t i m e o f f l i g h t = 32.5 + 3 = 35.5 s h o r i z o n t a l r a n g e = 114 \times 32.5 + 261 = 3966 m$

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