Question

Long Jump: inital center of mass height of 1.08 m, final center of mass height of 0.42 m, projection velocity of 8.7 m/s, projection angle of 22.3 degrees, and leg extension of 77 m.

1) How long does it take to reach maximum jump height? A. 0.337 s B. 0.674 s C. 0.821 s D. 0.887 s

2) What is the maximum center of mass height off of the ground? A. 1.08 m B. 1.64 m C. 4.38 m D. 4.94 m

3) What is the time of flight? A. 0.586 s B. 0.674 s C. 0.834 s D. 0.915 s

4) What is the distance of jump? A. 6.20 m B. 7.49 m C. 8.03 m D. 8.73 m

Answer 1

1) The maximum jump height is reached at A.
`0.337s`

2) The maximum center of mass height off of the ground is B.
`1.64m`

3) The time of flight is C.
`0.834s`

4) The distance of jump is B.
`7.49m`

Step-by-step 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`

1) We use,
`v_(fy)=v_(iy)-gt`, as we clear it for
`t` and using the fact that
`v_(fy)=0` at max height, we obtain
`t=(v_(iy))/(g) =(3.3m/s)/(9,8m/s^(2)) =0.337s`

2) We can use the formula
`y_(max)=y_(i)+v_(iy)t-(gt^(2))/(2)` for
`t=0.337s`, so

`y_(max)=1.08m+(3.3m/s)(0.337s)-((9.8m/s^(2))(0.337)^(2))/(2)=1.64m`

3) We can use the formula
`y_(f)=y_(i)+v_(iy)t-(gt^(2))/(2)`, to find total time of fligth, so
`0.42=1.08+3.3t-((9.8)t^(2))/(2)\\0=-4.9t^(2)+3.3t+0.66`, as it is a second-grade polynomial, we find that its positive root is
`t=0.834s`

4) Finally, we use
`x=v_(x)t=8.05m/s(0.834s)=6.71m`, as it has an additional displacement of
`0.77m` due the leg extension we obtain,

`x=6.71m+0.77m=7.48m`, aprox
`x=7.49m`