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Question

Two bodies 1 and 2 are projected simultaneously with velocities $v_{1} = 2 {m s}^{- 1}$ and $v_{2} = 4 {m s}^{- 1}$ . The body 1 is projected vertically up from the top of a cliff of height $h = 10 m$ and the body 2 is projected vertically up from the bottom of the cliff. If the bodies meet, find the time (in $s$ ) of meeting of the bodies.

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Solution

We will start with writing the equation of motion for both bodies in vertical direction:
$B o d y 1 :$ $s_{1} = 10 + 2 t - \frac{g t^{2}}{2}$
$B o d y 2 :$ $s_{2} = 4 t - \frac{g t^{2}}{2}$
when they meet $s_{1} = s_{2}$
hence, $10 + 2 t = 4 t$ or $t = 5 s$

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We will start with writing the equation of motion for both bodies in vertical direction:
$B o d y 1 :$ $s_{1} = 10 + 2 t - \frac{g t^{2}}{2}$
$B o d y 2 :$ $s_{2} = 4 t - \frac{g t^{2}}{2}$
when they meet $s_{1} = s_{2}$
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