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We need to find the velocity of the glider just after the skydiver lets go .

When the initial momentum equals the final momentum, the momentum of an isolated system changes to zero. The particle interactions are isolated from the external environment. Because momentum is conserved, we apply the law of conservation of momentum, which is given by equation in the form

${\stackrel{鈫�}{P}}_f={\stackrel{鈫�}{P}}_i\left(1\right)$

Because of isolated system the railway is frictionless. The object has mass $m$and moves with speed $v$that has momentum $p$, Vector is a product of object's mass and its velocity.The momentum is given by equation in the form

$p=mv\left(2\right)$

Glider's initial velocity is $v_i=30m/s$and total initial mass is $m_i=680kg$. To get initial momentum ${\stackrel{鈫�}{p}}_i$

$$\begin{gathered} p_i=m_i{v}_i \\ =\left(680kg\right)\left(30m/s\right) \\ =(20400kg)脳(m/s) \end{gathered}$$

The final mass of glider after skydiver drops out is

$m_f=680kg-60kg=620kg$

Skydivers gain the velocity of gliders, so our final system is a combination of gliders and skydivers. To calculate the final momentum,

$$\begin{gathered} p_f=m_f{v}_f+m_{skydiver}{v}_{skydiver} \\ =\left(620kg\right)v_f+\left(60kg\right)\left(30m/s\right) \\ =\left(620kg\right)v_f+1800 \end{gathered}$$

To get the equation for $v_f$, put the expression into equation $\left(1\right)$

$$\begin{gathered} p_i=\left(620kg\right)v_f+1800 \\ {v}_f=\frac{p_i-1800}{620kg}\left(3\right) \end{gathered}$$

After putting values for $p_i$into equation $\left(3\right)$to get $v_f$

$$\begin{gathered} v_f=\frac{p_i-1800}{620kg} \\ =\frac{(20400kg)脳(m/s)-1800}{620kg} \\ =30m/s \end{gathered}$$