91Ó°ÊÓ

The rock’s velocity is v=12 m/s at an angle above the horizontal.

Your mass is m =70.0 kg.

Mass of rock is $m\text{'}=3.0kg$.

The total momentum of all the constituents of an isolated system should remain unchanged during the collision process, though the momentum individual constituents may change.

Let the horizontal component of velocity of rock $\left(v_x\right)$is given as-

$$\begin{gathered} {\mathrm{v}}_{\mathrm{x}}=\mathrm{v}\cos 35° \\ =\left(12\mathrm{m}/\mathrm{s}\right)\cos 35° \\ =9.83\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the horizontal component of this velocity is 9.83 m/s.

The horizontal component of the rock’s momentum is:

$$\begin{gathered} p=m\text{'}{v}_x \\ =3\mathrm{kg}×9.83\mathrm{m}/\mathrm{s} \\ =29.5\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

The momentum of the person is calculated as follows:

$$\begin{gathered} p\text{'}=mv\text{'} \\ =\left(70kg\right)v\text{'} \end{gathered}$$

Now, apply law of conservation of momentum as follows:

$$\begin{gathered} p=p\text{'} \\ 29.5 kgâ‹\dots m/s=(70 kg)×v\text{'} \\ v\text{'}=\frac{29.5kg.m/s}{70kg} \\ v\text{'}=0.42\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed after throw of rock is 0.42 m/s.