# Jhjhj

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• Publicado : 12 de marzo de 2012

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3.13: Take +y to be upward.
Use the vertical motion to find the time in the air:

Then

b) since

3.21: a) The time the quarter is in the air is thehorizontal distance divided by the horizontal component of velocity. Using this time in Eq. (3.18),

b) Using the same expression for the time in terms of the horizontaldistance in
Eq. (3.17),

3.30: , corresponding to a period of 0.109 s. a) From Eq. (3.29), . b) From either Eq. (3.30) or Eq. (3.31), .
3.34: a) , and . So, tothe right of vertical.
b)

3.69: Take to be upward.
a) The vertical motion of the shell is unaffected by the horizontal motion of the tank. Use the verticalmotion of the shell to find the time the shell is in the air:

Relative to tank #1 the shell has a constant horizontal velocity . Relative to the ground the horizontalvelocity component is . Relative to tank #2 the shell has horizontal velocity component . The distance between the tanks when the shell was fired is the that the shelltravels relative to tank #2 during the 8.86 s that the shell is in the air.
b) The tanks are initially 2000 m apart. In 8.86 s tank #1 travels 133 m and tank #2 travels310 m, in the same direction. Therefore, their separation increases by . So, the separation becomes 2180 m (rounding to 3 significant figures).

3.77: a) The path is acycloid.

b) To find the velocity components, take the derivative of x and y with respect to time: and To find the acceleration components, take the derivative of andwith respect to time: and
c) The particle is at rest every period, namely at At that time, and The acceleration is in the direction.
d) No, since

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