# Question #9cedf

##### 1 Answer

#### Answer:

#### Explanation:

The idea here is that the ratio of the ball's velocity when it first makes contact with the ground, and its velocity when it leaves the ground **does not** depend on the value of

Let's take

The ball is dropping from

#v_1^2 = overbrace(color(white)(a)v_0^2color(white)(a))^(color(blue)(=0)) + 2 * g * h_1#

#v_1^2 = 2 * g * h_1" " " "color(orange)((1))#

After it hits the ground, it bounces with an initial velocity of

#overbrace(color(white)(a)v_f^2color(white)(a))^(color(blue)(=0)) = v_2^2 - 2 * g * h_2#

#v_2^2 = 2 * g * h_2" " " "color(orange)((2))#

Now, in order to find the factor by which the ball loses its velocity, divide equation

#v_2^2/v_1^2 = (color(red)(cancel(color(black)(2 * g))) * h_2)/(color(red)(cancel(color(black)(2 * g))) * h_1)#

This gets you

#v_2/v_1 = sqrt(h_2/h_1)#

Plug in your values to find

#v_2/v_1 = sqrt(( 1.8 color(red)(cancel(color(black)("m"))))/(5color(red)(cancel(color(black)("m"))))) = sqrt(36/100) = 6/10 = 3/5#

You can thus say that

#v_2 = 3/5 * v_1#

Sine the velocity of the ball **decreased** by

#v_1 - v_2 = v_1 - 3/5 * v_1 = (1 - 3/5) * v_1 = 2/5 * v_1#

you can say that it decreased by a **factor** of

#(v_1 - v_2)/v_1 = (2/5 * color(red)(cancel(color(black)(v_1))))/color(red)(cancel(color(black)(v_1))) = color(green)(|bar(ul(color(white)(a/a)color(black)(2/5)color(white)(a/a)|)))#