Answer
Apple 2 is thrown down with a speed of $~~9.56~m/s$
Work Step by Step
From the graph, we can see that it takes 2.0 seconds for apple 1 to reach the roadway.
We can find the height of the bridge:
$y = \frac{1}{2}at^2$
$y = \frac{1}{2}(9.8~m/s^2)(2.0~s)^2$
$y = 19.6~m$
Since $t_s = 2.0~s$, we can see that each block of time is $0.25~s$
Thus apple 2 reached the roadway in a time of $1.25~s$
We can find the initial speed of apple 2:
$y = v_0~t+\frac{1}{2}at^2$
$v_0~t = y - \frac{1}{2}at^2$
$v_0 = \frac{y}{t} - \frac{1}{2}at$
$v_0 = \frac{19.6~m}{1.25~s} - \frac{1}{2}(9.8~m/s^2)(1.25~s)$
$v_0 = 9.56~m/s$
Apple 2 is thrown down with a speed of $~~9.56~m/s$
