Very lost please help.
1 answer:
A) The acceleration is due to gravity at any given point if you look at it vertically, so 
.
b)
, so 
. We use 
and then the final speed must be 0 because it stops at the highest point. So 
. Solve for 
and you get 
c)
, and then we plug the values: 
and we already have the time from "b)", so ![Y_m_a_x = [(32sin(25))*(32sin(25)/10)] - 5(32sin(25)/10)^2](https://tex.z-dn.net/?f=Y_m_a_x%20%3D%20%5B%2832sin%2825%29%29%2A%2832sin%2825%29%2F10%29%5D%20-%205%2832sin%2825%29%2F10%29%5E2)
; then we just rearrange it ![Y_m_a_x = 10[(32sin(25))^2/100] - 5 [(32sin(25))^2/100]](https://tex.z-dn.net/?f=Y_m_a_x%20%3D%2010%5B%2832sin%2825%29%29%5E2%2F100%5D%20-%205%20%5B%2832sin%2825%29%29%5E2%2F100%5D%20)
and finally ![Y_m_a_x = 5[(32sin(25))^2/100] = (32sin(25))^2/20](https://tex.z-dn.net/?f=Y_m_a_x%20%3D%205%5B%2832sin%2825%29%29%5E2%2F100%5D%20%3D%20%2832sin%2825%29%29%5E2%2F20)
b)
c)
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Answer:
h = 3.56 m
Explanation:
- Assuming no friction on the rope or air resistance, we can apply the conservation of mechanical energy principle to the first part of the trajectory, from when Jane starts her swing till she catches Tarzan at the ground level.
⇒
- Rearranging terms, we get:
- Now, as Jane starts from rest, Kij =0.
- if we choose the ground level as our zero reference level, it will be also Ufj = 0.
- Replacing in (2) by the expressions of Uij and Kfj, we have:
- Replacing in (3) by the givens, and rearranging terms, we can solve for vfj, as follows:
- Now, as once jane catches Tarzan, both continue swinging together, we can take the catching moment as a completely inelastic collision.
- Assuming no external forces act during the collision, total momentum must be conserved.
⇒
- Assuming that Tarzan is at rest when Jane catches him, the initial momentum will be simply as follows:
- The final momentum, will be just the product of the combined mass of Jane and Tarzan times the common speed for them after the collision:
- As (6) and (7) are equal each other, we can solve for vjt, as follows:
- Finally we can apply the same energy conservation principle to the last part of the trajectory, as follows:
- We know that Uijt = 0 and also that Kfjt = 0, due to both starts from the ground level and reach to the highest point before starting to fall down, so at this point, the kinetic energy will be zero.
- Replacing by the givens and the result from (8), we can solve for the h as follows: