Answer:
a)
![V_H=0.056 (-i)m/s](https://img.qammunity.org/2020/formulas/physics/college/hv3ix1bz68mr8ov10fau31pttmw1eaeskm.png)
b)
![V_H=0.031 (-i)m/s](https://img.qammunity.org/2020/formulas/physics/college/fws4esrerzscu4hg2opafyfxii1ca2aj4t.png)
Step-by-step explanation:
By conservation of the linear momentum:
![P_o = P_f](https://img.qammunity.org/2020/formulas/physics/college/6m6nqc9i6g8vtcrfihe8arroyrhkwdiojn.png)
![0 = M_H*V_H+m_b*V_b](https://img.qammunity.org/2020/formulas/physics/college/ciyfl4fe0ptgxi3mdvcxx29pyyqpqf9q3l.png)
![V_H = m_b/M_H*V_b](https://img.qammunity.org/2020/formulas/physics/college/1bgk4i4w4ym1y09n9fe8a8dm8rapgtof4v.png)
When the bullet is shot horizontally
![V_b = 965m/s](https://img.qammunity.org/2020/formulas/physics/college/fs23iihocild0g9fjwzb9tzg9mcfry155r.png)
![V_H = -0.056m/s](https://img.qammunity.org/2020/formulas/physics/college/kd6gv7770cwn9dpuizrmsg6wwe7daa6kak.png)
When the bullet is shot at 56.0° above the horizontal,
![V_b = 965*[cos(56),sin(56)]m/s=[539.62;800.02]m/s](https://img.qammunity.org/2020/formulas/physics/college/dhcnps72wustbxi7zxi34uz439yw8abtb6.png)
Since on the y-axis ground will absorb the shock, we only make the calculations on the x-axis:
![V_H = -0.031m/s](https://img.qammunity.org/2020/formulas/physics/college/442pgubz4mr2xpog8dpcfcuevjx820wvk4.png)
![V_H = -0.056m/s](https://img.qammunity.org/2020/formulas/physics/college/kd6gv7770cwn9dpuizrmsg6wwe7daa6kak.png)