Two bodies are thrown simultaneously from a t | vedclass

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(B) Let the tower be the origin. For the body thrown vertically upwards,the displacement after time $t$ is $y_1 = v_0t - \frac{1}{2}gt^2$.For the body

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$2v_0t$

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(B) Let the tower be the origin. For the body thrown vertically upwards,the displacement after time $t$ is $y_1 = v_0t - \frac{1}{2}gt^2$.For the body thrown vertically downwards,the displacement after time $t$ is $y_2 = -v_0t - \frac{1}{2}gt^2$ (taking downward direction as negative).The distance between the two bodies is the magnitude of the difference in their positions: $d = |y_1 - y_2|$.$d = |(v_0t - \frac{1}{2}gt^2) - (-v_0t - \frac{1}{2}gt^2)|$.$d = |v_0t - \frac{1}{2}gt^2 + v_0t + \frac{1}{2}gt^2| = |2v_0t| = 2v_0t$.

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

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