What is the magnitude of the instantaneous velocity (speed) of a freely falling object 13 s after it is released from a position of rest? Consider the acceleration of gravity to be 10 m/s2 .
To find the magnitude of the instantaneous velocity of a freely falling object 13 seconds after it is released, we can use the formula for a constant acceleration motion:
v = u + at
Where:
v is the final velocity (unknown)
u is the initial velocity (0 m/s, since the object is at rest)
a is the acceleration due to gravity (-10 m/s^2, considering it is downward)
t is the time (13 seconds)
Substituting the given values into the formula, we have:
v = 0 + (-10 m/s^2) * 13 s
Now we can calculate the velocity:
v = -10 m/s^2 * 13 s
v = -130 m/s
The magnitude of the velocity represents the speed, so we ignore the negative sign and take the absolute value:
Magnitude of velocity = |v| = |-130 m/s| = 130 m/s
Therefore, the magnitude of the instantaneous velocity of the freely falling object 13 seconds after being released is 130 m/s.