Problem: One object moves horizontally at 57 m/s, while another falls freely to the ground under the influence of gravity, from rest, a distance of 67 meters. How long does it take the second object to reach the ground? How far does the first object travel from the time the second object starts falling until it strikes the ground?
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Solution: To find the distance of the first object we need only find the time, and multiply by its velocity. The time will be the time of fall of the second object, so we momentarily forget about the first object and determine how long the second object requires to fall. We use the formula `ds = v0 * `dt + .5 a `dt ^ 2, with v0=0, a=9.8 m/s/s and s= 67 m.
We obtain the equation 67 m = (0 m/s) *`dt + .5 (9.8 m/s/s) * `dt ^ 2, or 67 m = (4.9 m/s/s)`dt ^ 2.
Solving for `dt we obtain `dt = `sqrt{( 67 m/s)/4.9 m/s/s)} = 3.697 s.
The distance traveled by the first object in this time is thus ( 57 m/s)( 3.697 s) = 210.7 m.
.
.
.
.
.
.
.
.
.
.