91Ó°ÊÓ

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.

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

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