A particle-is projected vertically upwards with an initial velocity of 40 m/s. Find the displacement and distance covered by the particle in 6 seconds. Take g=10 m/s $^{2}$

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Solution

$T i m e o f f l i g h t = \frac{2 u}{g} = 8 s t h u s i s m e a n s t h a t t h e s t o n e i s m o v i n g d o w n w a r d s a f t e r 6 s d i p l a c e m e n t = u t + \frac{1}{2} a t^{2} = 40 \times 6 - \frac{1}{2} \times 10 \times 6^{2} = 60 d i t a n c e = H + (H - 60) = 2 H - 60, w h e r e H i s m a x h e i g h t H = \frac{u^{2}}{2 g} = \frac{1600}{20} = 80 m d i s t a n c e = 100 m$

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