Two bodies are projected vertically upwards from one point with the same initial velocity $v_{0} .$ The second body is projected $t_{0} s$ after the first. How long after will the bodies meet?

\frac{v_{0}}{2 g} + \frac{t_{0}}{2}

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\frac{v_{0}}{g} + \frac{t_{0}}{2}

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\frac{v_{0}}{g} + \frac{t_{0}}{4}

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\frac{2 v_{0}}{g} + \frac{t_{0}}{2}

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Solution

The correct option is B $\frac{v_{0}}{g} + \frac{t_{0}}{2}$

$t h e d i s p l a c e m e n t o f t h e t w o b o d i e s s h o u l d b e s a m e w h e n t h e y c o l l i d e h = v_{0} t - \frac{1}{2} g t^{2} = v_{0} (t - t_{0}) - \frac{1}{2} g {(t - t_{0})}^{2} \Rightarrow t = \frac{v_{0}}{g} + \frac{t_{0}}{2}$

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Two bodies are projected vertically upwards from one point with the same initial velocity v0. The second body is projected t0s after the first. How long after will the bodies meet?

The correct option is B v0g+t02 thedisplacementofthetwobodiesshouldbesamewhentheycollideh=v0t−12gt2=v0(t−t0)−12g(t−t0)2⇒t=v0g+t02

Two bodies are projected vertically upwards from one point with the same initial velocity $v_{0} .$ The second body is projected $t_{0} s$ after the first. How long after will the bodies meet?