Answer:
The answer is "[tex]4950 \frac{g\ cm }{s}[/tex]"
Explanation:
Consider vertical component of motion:
[tex]\to s_y=u_yt+ \frac{1}{2} at^2\\\\1.0315=0+\frac{1}{2} \times 9.81 \times t^2\\\\1.0315=\frac{1}{2} \times 9.81 \times t^2\\\\1.0315=4.905 \times t^2\\\\t^2= \frac{1.0315}{4.905}\\\\t=0.207059711\\\\[/tex]
Considering the horizontal components:
[tex]\to v_x=x\times t\\\\[/tex]
[tex]=03.6025 \times 0.207059711\\\\=0.745 \frac{m}{s}\\\\\to p=mv\\\\=66.45 \times0.745 \times 10^2\\\\=4950\ \frac{g\ cm }{s}[/tex]
