A body is projected vertically downward from a top of tower of h=40m with velocity =5m/s Another body is projected up from the bottom of tower with vertically upward velocity =15m/s .When &where will bodies meet?

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Solution

$L e t t h e b o d i e s m e e t a f t e r t s e c o n d s . S i g n c o n v e n t i o n : U p : + v e a n d D o w n : - v e D i s \tan c e t r a v e l l e d b y v e r t i c a l l y u p w a r d p r o j e c t e d b o d y : s_{1} = u t - \frac{1}{2} g t^{2} = 15 t - \frac{1}{2} \times 9.8 t^{2} = 15 t - 4.9 t^{2} D i s t a n c e t r a v e l l e d b y v e r t i c a l l y d o w n w a r d p r o j e c t e d b o d y : s_{2} = u t + \frac{1}{2} g t^{2} = 5 t + \frac{1}{2} \times 9.8 t^{2} = 5 t + 4.9 t^{2} s_{1} + s_{2} = 40 20 t = 40 t = 2 s D i s \tan c e o f m e e t i n g p o i n t f r o m t h e g r o u n d : s_{1} = 15 t - 4.9 t^{2} = 15 \times 2 - (4.9 \times 4) = 10.4 m$

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