Question

the stone is allowed to fall from the top of the tower 100m high and at the same time another stone is projected vertically upward from the ground with a velocity of 25 m/s. calculate when and where does two stones will meet​

Answer 1

Explanation:

Hope it helps

Ask if any confusion

Answer 2

Answer:

The two stones will meet after 4 seconds when the falling stone has covered a height of 80m.

Explanation:

Height ( h )       =    100m

Time ( t )          =   ?

 g                   =  10 m/s²  

Height covered by the falling stone = s₁

Therefore,      s₁  =  ut + 1/2gt²

                       s₁  =  0 × t + 1/2 ( 10 ) t²

                       s₁ =  5t²                              .....................................   ( 1 )

Let ,The distance covered by the stone thrown upward = s₂

                         g = ₋ 10 m/s

                        u  =  25 m/s

                       s₂  = ut + 1/2 gt²

Therefore,       s₂  =  25t + 1/2 ( - 10 ) t²

                        s₂  =  25t -5t²               .....................................  ( 2 )

Total height given = 100m

∴           s₁ + s₂     =  100m

5t² + ( 25t - 5t² )  =  100m

∴   25t                  =  100m

       t                  = 100/25 = 4 seconds  

                                                                      ..................................... ( 3 )

Putting the value of equation ( 3 ) in equation ( 1 ), we get

∴       s₁    =   5 t²

               =   5 × ( 4 )²

               =  80m

Therefore, The two stones will meet after 4 seconds when the falling stone has covered a height of 80m.