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Set 4 Problem number 18
A projectile is fired in a horizontal direction at 11 meters/second. It continues
traveling horizontally at this rate.
- How far does it travel in the horizontal direction during a vertical free
fall through 31 meters?
The motion of this object is equivalent to the motion of two objects, one moving
horizontally and one vertically, with the specified distances, speeds and accelerations.
To find the distance traveled by the first object we need only find the
time of fall, and multiply by its constant horizontal velocity.
- The time will be the time of fall in the vertical direction, so we
momentarily forget about the horizontal direction and determine the time duration of the
vertical motion.
- To find this time we can use the formula `dsy = v0y * `dt + .5 ay `dt ^
2, with v0y=0, ay=9.8 m/s/s and `dsy= 31 m.
- We obtain `ds = .5 a `dt^2, since v0 = 0.
- Solving for `dt and rejecting the negative solution we get `dt = `sqrt( 2
* `ds / a ).
- Substituting we obtain `dt = `sqrt{( 31 m/s)/4.9 m/s/s)} = 2.515 s.
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- The horizontal displacement in this time is
- horizontal displacement = `dsx - ( 11 m/s)( 2.515 s) = 27.66 m.
When the only force acting on an object is a uniform gravitational acceleration
in the vertical, or y, direction, the object's horizontal (x) and vertical (y) motions are
completely independent.
- Since the net force is in the vertical direction, the acceleration in the y
direction with be that of gravity.
- Since there is no net force in the horizontal direction, the acceleration in the
x direction will be zero.
The object will fall freely a distance `dsy from rest in time `dt = `sqrt( 2
`dsy / g ), obtained from `dsy = v0y * `dt + .5 ay `dt^2 with v0 = 0 and ay = g.
In this time the horizontal displacement will be
- `dsx = vx * `dt = vx * `sqrt( 2 * `dsy / g).
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