"
Set 4 Problem number 17
An object moves both horizontally and
vertically.
Its horizontal motion is at a constant velocity
of 14 m/s; its vertical motion begins at velocity zero and it accelerates in that
direction at a constant rate of 9.8 m/s/s.
- How far does it travel horizontally, and how far
vertically, in 3.4 seconds?
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 3.4 seconds the first object, traveling at 14 m/s, will travel ( 14
m/s)( 3.4 sec) = 47.6 meters.
- The second object will attain a speed of (9.8 m/s/s)( 3.4 sec) = 33.32 m/s;
- its average speed will be (0+ 33.32 m/s)/2 = 16.66 m/s, and
- the distance will be ( 16.66 m/s)( 3.4 sec) = 56.64 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.
"