Question

A ball is thrown vertically upward with a speed u it reaches a point B at a height h after time t1 it returns to the ground after time T2 from the instant it was at B during the upward journey then,t1t2is equal to

Answer 1

Answer:

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Answer 2

Question

      A ball is thrown vertically upward with a speed u it reaches a point B at a height h after time t1 it returns to the ground after time T2 from the instant it was at B during the upward journey then,t1t2is equal to

Answer:

At point B : the velocity of ball is

    $v = u - gt_{1}$   ------- (1)

the height of point B is given as

   $h = ut_{1} - \frac{1}{2} * g t^2$ -------- (2)

from point B the ball takes time $t_{2}$ to reach to its intial position .

∴ $-h = vt_{2} - \frac{1}{2} gt_{2} ^2$  -------- (3)

now ,

        substracting eqn.

         (2) with eqn. (3)

we get ,

       $2h = ( u t_{1} - ( u - gt_{1} )t_{2} ) - \frac{1}{2} g( t_{1} - t_{2} )$

from eqn. (1)

      $2h = ( u t_{1} - ( u - gt_{1} )t_{2} ) - \frac{1}{2} g( t_{1} - t_{2} )$

      $2h = u t_{1} - u t_{2} + g t_{1} t_{2} - \frac{1}{2} g( t_{1} - t_{2} ) ( t_{1} + t_{2} )$

      $now t_{1} + t_{2} = time of flight = \frac{2u}{g}$

      $2h = u( t_{1} - t_{2} ) + g t_{1} t_{2} - \frac{1}{2} g( t_{1} - t_{2} ) * \frac{2u}{g}$

      $2h = g x_{1} x_{2}$

    ∴$t_{1} t_{2} = \frac{2u}{g}$

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