# Question 93a60

Sep 6, 2015

The answer is (b) ${\text{2100 ms}}^{- 2}$

#### 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

${v}_{\text{before"^2 = v_o^2 + 2 * g * h" }}$, where

${v}_{0}$ - the initial velocity of the ball - presumably this is equal to zero;
$h$ - the height from which it falls.

Plug in your values to get

${v}_{\text{before" = sqrt(2 * 9.8"m"/"s"^2 * "10 m") = sqrt(196"m"^2/"s"^2) = "14 m/s}}$

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 ${v}_{\text{top}}$ is equal to zero because the ball stops at the top of its motion, right before falling for a second time.

So, the velocity right after impact will be

${v}_{\text{after" = sqrt(2 * 2 * 9.8"m"/"s"^2 * "2.5 m") = sqrt(196"m"^2/"s"^2) = "7 m/s}}$

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 $+ \text{14 m/s}$ to $- \text{7 m/s}$, because in the same refrence frame the velocity after the impact is oriented upwards.

The average acceleration during contact will thus be

$a = \frac{{v}_{\text{before" - v_"after}}}{t}$

$a = \left({\text{14 m/s" - (-"7 m/s"))/"0.01 s" = 21/0.01"m"/"s"^2 = color(green)("2100 ms}}^{- 2}\right)$