Answer:
The displacement is [tex]\Delta H = - 1 \ m[/tex]
The distance is [tex]D = 4 \ m[/tex]
Explanation:
From the question we are told that
The height from which the ball is dropped is [tex]h = 1 \ m[/tex]
The height attained at the first bounce is [tex]h_1 = 0.8 \ m[/tex]
The height attained at the second bounce is [tex]h_2 = 0.5 \ m[/tex]
The height attained at the third bounce is [tex]h_3 = 0.2 \ m[/tex]
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
[tex]\Delta H = (0 - h ) + ( h_1 - h_1 ) + (h_2 - h_2 )+ (h_3 - h_3)[/tex]
Here the 0 show that there was no bounce back to the point where Billy released the ball
[tex]\Delta H = (0 - 1 ) + ( 0.8- 0.8 ) + (0.5 - 0.5 )+ (0.2 - 0.2)[/tex]
=> [tex]\Delta H = - 1 \ m[/tex]
Generally the distance covered by the ball is mathematically represented as
[tex]D = h + 2h_2 + 2h_3 + 2h_3[/tex]
The 2 shows that the ball traveled the height two times
[tex]D = 1 + 2* 0.8 + 2* 0.5 + 2* 0.2[/tex]
=> [tex]D = 4 \ m[/tex]
