Two bodies are thrown simultaneously from a tower with same initial velocity {v_0}: one vertically u

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2.Motion in Straight Line

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Two bodies are thrown simultaneously from a tower with same initial velocity ${v_0}:$ one vertically upwards, the other vertically downwards. The distance between the two bodies after time $t$ is

A

$2{v_0}t + \frac{1}{2}g\,{t^2}$

B

$2{v_0}t$

C

${v_0}t + \frac{1}{2}g\,{t^2}$

D

${v_0}t$

Solution

(b) For vertically upward motion, ${h_1} = {v_0}t – \frac{1}{2}g\,{t^2}$and for vertically down ward motion, ${h_2} = {v_0}t + \frac{1}{2}g\;{t^2}$

$\therefore $ Total distance covered in $t \,sec$ $h = {h_1} + {h_2} = 2{v_o}t$.

Standard 11

Physics

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