An object falls freely from rest for 11.25 meters before striking the ground. Another object starts at the same time from the same location and travels at 55.1625 m/s.
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 = 11.25 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{( 11.25 m/s)/4.9 m/s/s)} = 1.515 s.
- `dsx = ( 55.1625 m/s)( 1.515 s) = 83.57119 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