Answer:
1) The maximum jump height is reached at A. ![0.337s](https://tex.z-dn.net/?f=0.337s)
2) The maximum center of mass height off of the ground is B. ![1.64m](https://tex.z-dn.net/?f=1.64m)
3) The time of flight is C. ![0.834s](https://tex.z-dn.net/?f=0.834s)
4) The distance of jump is B. ![7.49m](https://tex.z-dn.net/?f=7.49m)
Explanation:
First of all we need to decompose velocity in its rectangular components, so
![v_{xi}=8.7m/s(cos 22.3\°)=8.05m/s= constant\\v_{yi}=8.7m/s(sin 22.3\°)=3.3m/s](https://tex.z-dn.net/?f=v_%7Bxi%7D%3D8.7m%2Fs%28cos%2022.3%5C%C2%B0%29%3D8.05m%2Fs%3D%20constant%5C%5Cv_%7Byi%7D%3D8.7m%2Fs%28sin%2022.3%5C%C2%B0%29%3D3.3m%2Fs)
1) We use,
, as we clear it for
and using the fact that
at max height, we obtain ![t=\frac{v_{iy}}{g} =\frac{3.3m/s}{9,8m/s^{2}} =0.337s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bv_%7Biy%7D%7D%7Bg%7D%20%3D%5Cfrac%7B3.3m%2Fs%7D%7B9%2C8m%2Fs%5E%7B2%7D%7D%20%3D0.337s)
2) We can use the formula
for
, so
![y_{max}=1.08m+(3.3m/s)(0.337s)-\frac{(9.8m/s^{2})(0.337)^{2}}{2}=1.64m](https://tex.z-dn.net/?f=y_%7Bmax%7D%3D1.08m%2B%283.3m%2Fs%29%280.337s%29-%5Cfrac%7B%289.8m%2Fs%5E%7B2%7D%29%280.337%29%5E%7B2%7D%7D%7B2%7D%3D1.64m)
3) We can use the formula
, to find total time of fligth, so
, as it is a second-grade polynomial, we find that its positive root is
4) Finally, we use
, as it has an additional displacement of
due the leg extension we obtain,
, aprox ![x=7.49m](https://tex.z-dn.net/?f=x%3D7.49m)