The two cylindrical rod segments are fixed to the rigid walls such that there is a gap of 0.01 in. between them when t1 = 60°f.
2 answers:
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The time taken by the ballast bag to reach the ground is 2.18 s
The ballast bag at rest with respect to the balloon has the upward velocity (u) of 4.6 m/s , which is the velocity of the balloon. When it is dropped from the balloon, its motion is similar to an object thrown upwards with an initial velocity <em>u </em>and it falls under the acceleration due to gravity<em> g.</em>
Taking the upward direction as positive and the downward direction as negative, the following equation of motion may be used.
The bag makes a net displacement <em>s</em> of 13.4 m downwards, hence
Its initial velocity is
The acceleration due to gravity acts downwards and hence it is negative.
Use the values in the equation of motion and write an equation for t.
Solving the equation for t and taking only the positive value for t,
t=2.18 s
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: