Question

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

bounces again to a height of 0.5 m, and bounces once more to a height of 0.2 m before

coming to a rest. Up is the positive direction. What are the total displacement of the ball and

the total distance traveled by the ball?

Answer 1

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

The distance is
`D = 4 \ m`

Step-by-step 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`