# Question #93a60

##### 1 Answer

#### Answer:

The answer is **(b)**

#### Explanation:

The idea here is that the acceleration of the ball during contact will be determined by how its velocity changes from **right before impact** to **immediately after impact**.

Since it's falling from a height of **10 m**, its velocity right before impact will be

#v_"before"^2 = v_o^2 + 2 * g * h" "# , where

Plug in your values to get

#v_"before" = sqrt(2 * 9.8"m"/"s"^2 * "10 m") = sqrt(196"m"^2/"s"^2) = "14 m/s"#

**After impact**, the ball reaches a height of **2.5 m**, which means that it started moving up again with a velocity of

#overbrace(v_"top"^2)^(color(blue)(=0)) = v_"after"^2 - 2 * g * h_2#

Here

So, the velocity right after impact will be

#v_"after" = sqrt(2 * 2 * 9.8"m"/"s"^2 * "2.5 m") = sqrt(196"m"^2/"s"^2) = "7 m/s"#

Now, to get the total change in velocity, you need to consider the diraction of movement. Let's take the downward motion as being positive.

This means that the velocity of the ball went from *after the impact* is oriented **upwards**.

The average acceleration during contact will thus be

#a = (v_"before" - v_"after")/t#

#a = ("14 m/s" - (-"7 m/s"))/"0.01 s" = 21/0.01"m"/"s"^2 = color(green)("2100 ms"^(-2))#