Question

Billy drops a ball from a height of 1 m. The ball bounces back to a height of 0.8 m, then

Answer 1

Answer:

The displacement is  $\Delta H = - 1 \ m$

The distance  is  $D = 4 \ m$

Explanation:

From  the question we are told that

    The height from which the ball is dropped is  $h = 1 \ m$

    The height attained at  the first bounce is  $h_1 = 0.8 \ m$

    The height attained at  the second bounce is   $h_2 = 0.5 \ m$

    The height attained at  the third bounce is $h_3 = 0.2 \ m$

Note  : When calculating displacement we consider the direction of motion

Generally given that upward is positive  the total displacement of the ball is mathematically represented as

            $\Delta H = (0 - h ) + ( h_1 - h_1 ) + (h_2 - h_2 )+ (h_3 - h_3)$

Here the 0 show that there was no bounce back to the point where Billy released the ball  

           $\Delta H = (0 - 1 ) + ( 0.8- 0.8 ) + (0.5 - 0.5 )+ (0.2 - 0.2)$

=>          $\Delta H = - 1 \ m$

Generally the distance covered by the ball is mathematically represented as  

                $D = h + 2h_2 + 2h_3 + 2h_3$

The 2 shows that the ball traveled the height two times

              $D = 1 + 2* 0.8 + 2* 0.5 + 2* 0.2$

=>           $D = 4 \ m$