Answer
The second piton is thrown down with a speed of $~~17.25~m/s$
Work Step by Step
From the graph, we can see that it takes 3.0 seconds for first piton to fall.
We can find the distance the first piton falls in 3.0 seconds:
$y = \frac{1}{2}at^2$
$y = \frac{1}{2}(9.8~m/s^2)(3.0~s)^2$
$y = 44.1~m$
From the graph, we can see that the second piton fell 54.1 meters in 2.0 seconds.
We can find the initial speed of the second piton:
$y = v_0~t+\frac{1}{2}at^2$
$v_0~t = y - \frac{1}{2}at^2$
$v_0 = \frac{y}{t} - \frac{1}{2}at$
$v_0 = \frac{54.1~m}{2.0~s} - \frac{1}{2}(9.8~m/s^2)(2.0~s)$
$v_0 = 17.25~m/s$
The second piton is thrown down with a speed of $~~17.25~m/s$
