Question

- (11: A balloon is rising vertically upwards with uniform acceleration
15.7 m/8². A stone is dropped from it. After & hec another
stone is
dropped from it. Find the distance between the
two stones 6 seconds
after the second
stone is dropped by​

Answer 1

Answer:

h(10) - h(6) = distance between stones

h(10) = 9/2 (10^2) +15.7 (10^2)

h (6) = 9/2 (6^2) + 15.7 ( 6^2)

h(10) - h(6) = (15.7 - 4.9) (100) - (15.7 -4.9) (36)

= 1080 - 388.8

= 691 .2

*mark brainly*

Answer 2

GIVEN:

        Uniform acceleration of a balloon is $15.7m/s^{2}$

            A stone is dropped from it.

  TO FIND :

      The distance between the two stones 6 seconds after the second stone is dropped ...

      SOLUTION:

                     $S=u+\frac{1}{2} at^{2}$

                            AT initial point the u will be zero.

                    $a=15.7m/s^{2}$

                 First stone is dropped at zero sec.  

           Second stone is dropped after $6sec.$

        Then the first is dropped at $10sec$

         Distance between the stones is given by $h(10)-h(6)$

                      $h(10)=g/2(10^{2} )+15.7(10^{2})$

                      $h(6)=g/2(6^{2} )+15.7(6^{2} )$

                      $h(10)-h(6)=$$(15.7-4.9)(100)-(15.7-4.9)(36)$

                                        $=1080-388.8$

                                             $=691.2m$

         HENCE

                     The distance between the two stones 6 seconds after the second stone is dropped is given by   $691.2m$