Answer to Question #105463 in Mechanics | Relativity for havefun7741

Question

[img]https://upload.cc/i1/2020/03/13/uVG8F5.jpg[/img]


d8 is 4

Answer 1

(i)

$h_{max}=\frac{v_0^2\sin^2\alpha}{2g}=\frac{30^2\cdot\sin^250°}{2\cdot9.81}=35.14$ $m$

(ii)

$L=l_1+l_2=\frac{v_0^2\sin2\alpha}{g}+v_0\cos\alpha\cdot t=$

$=\frac{v_0^2\sin2\alpha}{g}+v_0\cos\alpha\cdot \frac{\sqrt{2gh+v_0^2}-v_0}{g}=$

$=\frac{30^2\sin100°}{9.81}+30\cos50°\cdot \frac{\sqrt{2\cdot 9.81\cdot 30+(30\cdot\sin50°)^2}-30\cdot\sin50°}{9.81}=$

$=90.35+20.48=110.83$ $m$

(iii)

$v_x=v_0\cos50°=30\cdot\cos50°=19.28 m/s$

$v_y=\sqrt{2gh+(v_0\sin50°)^2}=\sqrt{2\cdot 9.81\cdot 30+(30\cdot\sin50°)^2}=$

$=33.42m/s$

$\overrightarrow{v}=19.28\overrightarrow{i}+33.42\overrightarrow{j}$

$v=\sqrt{19.28^2+33.42^2}=38.58m/s$

$\tan\alpha=\frac{v_x}{v_y}=\frac{19.28}{33.42}=0.5769 \to\alpha\approx30°$