a box is dropped off a tall building. At what time after release is it falling 15 m/s?
When g*t = 15 m/s.
Solve for t.
g is the acceleration of gravity.
15
To find the time at which the box is falling at a velocity of 15 m/s, we can use the equation of motion:
v = u + at
where:
- v is the final velocity (15 m/s in this case),
- u is the initial velocity (0 m/s as the box is dropped),
- a is the acceleration due to gravity (-9.8 m/s², since the box is falling downwards),
- and t is the time.
Rearranging the equation, we have:
t = (v - u) / a
Substituting the known values:
t = (15 m/s - 0 m/s) / -9.8 m/s²
t ≈ -1.53 seconds
The negative sign indicates that the box is falling downwards, which is the expected result. Therefore, the box reaches a velocity of 15 m/s after approximately 1.53 seconds of being dropped.