A lift starts moving upwards from ground with a constant acceleration ‘a cm/s 2 ’ . After t seconds a piece of stone is dropped from it. Show that the stone touches the ground after [t (a ½) ]/[a½ + (a+g) 1/2 ] seconds.

Question

Saurabh Koranglekar · Answer

Dear student v = at; s= 0.5at^2 stone droped v_up = at; a=-g distance to cover is =s= 0.5at^2 Now simplifying the equations we can find the required term Regards

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A lift starts moving upwards from ground with a constant acceleration ‘a cm/s 2 ’ . After t seconds a piece of stone is dropped from it. Show that the stone touches the ground after [t (a ½) ]/[a½ + (a+g) 1/2 ] seconds.

A lift starts moving upwards from ground with a constant acceleration ‘a cm/s²’ . After t seconds a piece of stone is dropped from it. Show that the stone touches the ground after [t (a^½)]/[a½ + (a+g)^1/2 ] seconds.

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A lift starts moving upwards from ground with a constant acceleration ‘a cm/s 2 ’ . After t seconds a piece of stone is dropped from it. Show that the stone touches the ground after [t (a ½) ]/[a½ + (a+g) 1/2 ] seconds.

A lift starts moving upwards from ground with a constant acceleration ‘a cm/s2’ . After t seconds a piece of stone is dropped from it. Show that the stone touches the ground after [t (a½) ]/[a½ + (a+g)1/2 ] seconds.

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A lift starts moving upwards from ground with a constant acceleration ‘a cm/s²’ . After t seconds a piece of stone is dropped from it. Show that the stone touches the ground after [t (a^½)]/[a½ + (a+g)^1/2 ] seconds.