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  • A stone dropped from a building of height h and it reaches after t seconds on earth. From the same building if two stones are thrown (one upwards and other downwards) with the same velocity u and t

A stone dropped from a building of height h and it reaches after t seconds on earth. From the same building if two stones are thrown (one upwards and other downwards) with the same velocity u and they reach the earth surface after 1 t and 2 t seconds respectively, then -

Option: 1

t=t_{1}-t_{2}
 


Option: 2

t=\frac{t_{1}+t_{2}}{2}

 


Option: 3

t=\sqrt{t_{1}t_{2}}


Option: 4

None of these


Answers (1)

best_answer

 

             1) If a body dropped from some height (initial velocity zero)

                      u = 0 

                      a = g

                     \begin{array}{l}{v=g t} \\ {h=\frac{1}{2} g t^{2}} \\ {v^{2}=2 g h} \\ {h_{n}=\frac{g}{2}(2 n-1)}\end{array}

       

For first case of dropping h= \frac{1}{2}gt^{2}

For sec case of downward throwing h=- ut_{1}+\frac{1}{2}gt^{2}_1=\frac{1}{2}gt^{2}

\Rightarrow - ut_{1}=\frac{1}{2}g\left ( t^{2} -t_{1}^{2}\right )

For sec case of upward throwing h=- ut_{2}+\frac{1}{2}gt^{2}_2=\frac{1}{2}gt^{2}

\Rightarrow - ut_{2}=\frac{1}{2}g\left ( t^{2} -t_{2}^{2}\right )

On solving these two equations: -\frac{t_{1}}{t_{2}}= \frac{t^{2}-t_{1}^{2}}{t^{2}-t_{2}^{2}}\Rightarrow t= \sqrt{t_{1}t_{2}}

 

Posted by

Rishi

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