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Set 4 Problem number 17
A projectile is fired in a horizontal direction at 18 meters/second. It continues
traveling horizontally at this rate.
- How far does it travel in the horizontal direction during the 2.6 seconds
required to reach the ground?
- Assuming that it undergoes free fall in the vertical direction, how far
does it fall vertically during this time?
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. Imagine the two objects.
- In 2.6 seconds the first object, traveling at 18 m/s, will travel ( 18
m/s)( 2.6 sec) = 46.8 meters.
- The second object will attain a speed of (9.8 m/s/s)( 2.6 sec) = 25.48 m/s;
- its average speed will be (0+ 25.48 m/s)/2 = 12.74 m/s, and
- the distance will be ( 12.74 m/s)( 2.6 sec) = 33.12 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.
- In time `dt, starting from rest, the object's vertical velocity will change by
`dvx = g * `dt (g is the acceleration of gravity).
- If the object starts with an initial vertical velocity vy0 = 0, then after
`dt seconds its vertical velocity will be vyf = g * `dt.
- Since the x acceleration is zero the x velocity vx will be constant.
- In time `dt the x displacement will therefore be `dsx = vx * `dt.
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