At a certain instant an object is released from rest into free fall, and a second object is struck, giving it a speed of 67.7875 meters/second in the horizontal direction. The second object travels in a straight line at this speed.
After falling precisely 12.5 meters, the first object strikes the second object.
To find the distance moved by the first object we need only find the time, and multiply by its velocity.
- We will use `dsy, v0y, vfy and ay for the displacement, initial and final velocities and acceleration in the vertical, or y, direction; `dsx, v0x, vfx and ax will stand for the corresponding quantities in the horizontal, or x, direction.
- To find `dt from the known vertical quantities we use the formula `dsy = v0 * `dt + .5 ay `dt ^ 2, with v0=0, ay=9.8 m/s/s and `dsy = 12.5 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{( 12.5 m/s)/4.9 m/s/s)} = 1.597 s.
- `dsx = ( 67.7875 m/s)( 1.597 s) = 108.2566 m.
An 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 an object moving at a constant horizontal velocity vx will travel distance