Answer:
1) The maximum jump height is reached at A. [tex]0.337s[/tex]
2) The maximum center of mass height off of the ground is B. [tex]1.64m[/tex]
3) The time of flight is C. [tex]0.834s[/tex]
4) The distance of jump is B. [tex]7.49m[/tex]
Explanation:
First of all we need to decompose velocity in its rectangular components, so
[tex]v_{xi}=8.7m/s(cos 22.3\°)=8.05m/s= constant\\v_{yi}=8.7m/s(sin 22.3\°)=3.3m/s[/tex]
1) We use, [tex]v_{fy}=v_{iy}-gt[/tex], as we clear it for [tex]t[/tex] and using the fact that [tex]v_{fy}=0[/tex] at max height, we obtain [tex]t=\frac{v_{iy}}{g} =\frac{3.3m/s}{9,8m/s^{2}} =0.337s[/tex]
2) We can use the formula [tex]y_{max}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}[/tex] for [tex]t=0.337s[/tex], so
[tex]y_{max}=1.08m+(3.3m/s)(0.337s)-\frac{(9.8m/s^{2})(0.337)^{2}}{2}=1.64m[/tex]
3) We can use the formula [tex]y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}[/tex], to find total time of fligth, so [tex]0.42=1.08+3.3t-\frac{(9.8)t^{2}}{2}\\0=-4.9t^{2}+3.3t+0.66[/tex], as it is a second-grade polynomial, we find that its positive root is [tex]t=0.834s[/tex]
4) Finally, we use [tex]x=v_{x}t=8.05m/s(0.834s)=6.71m[/tex], as it has an additional displacement of [tex]0.77m[/tex] due the leg extension we obtain,
[tex]x=6.71m+0.77m=7.48m[/tex], aprox [tex]x=7.49m[/tex]
