At time t = 0, a rock climber accidentally allows a piton to fall freely from a high point on the rock wall to the valley below him. Then, after a short delay, his climbing partner, who is 10higher on the wall, throws a piton downward. The positions yof the pitons versus tduring the falling are given in Fig. 2-43. With what speed is the second piton thrown?

Kinematics is the study of how a system of bodies moves without taking into account the forces or potential fields that influence motion. The equations which are used in the study are known as kinematic equations of motion. The kinematic equations relate the displacement, velocity, acceleration, and time to each other.

According to the kinematic equations,

$Y=\left(V_{jt}\right)\left(t\right)+\frac{1}{2}a{t}^2$

In case of a first piton, at t = 3 s, and initial velocity $V_i=0$. Therefore,

$$\begin{gathered} {\mathrm{y}}_{01}=-\frac{1}{2}(9.8)3^2 \\ =-44.1\mathrm{m} \end{gathered}$$

Sign is negative since displacement was taken in the downward direction. So, height would be 44.1 m.

For second piton, initial position of the piston,

localid="1656151763529" $$\begin{gathered} \mathrm{y}={\mathrm{y}}_{01}+10 \\ =44.1+10 \\ =54.1\mathrm{m} \\ {\mathrm{y}}_2=0\mathrm{at}t=3\sec \end{gathered}$$

According to the newton’s second kinematic equation,

$y_2-y=\left(V_{i2}\right)\left(t\right)+\frac{1}{2}a{t}^2$

In this case, the second piton was thrown at t = 1 s.

$$\begin{gathered} {\mathrm{y}}_2-{\mathrm{y}}_0=\left({\mathrm{V}}_{\mathrm{i}2}\right)(\mathrm{t}-1)+\frac{1}{2}(9.8){(\mathrm{t}-1)}^2-54.1=2\left({\mathrm{V}}_{\mathrm{i}2}\right)-19.6 \\ {\mathrm{V}}_{\mathrm{i}2}=17.25\mathrm{m}/\mathrm{s} \\ \approx 17\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of the second piton will be 17 m/s.