To find the instantaneous velocity of a freely falling object 10 seconds after it is released from rest, we can use the equation for constant acceleration:
v = u + at
Where:
v = final velocity
u = initial velocity (which is zero since the object is released from rest)
a = acceleration due to gravity (approximately 9.8 m/s², assuming no air resistance)
t = time (in this case, 10 seconds)
Substituting the known values into the equation, we get:
v = 0 + (9.8 m/s²)(10 s)
v = 98 m/s
Therefore, the instantaneous velocity of the object after 10 seconds is 98 m/s downward.
For the average velocity during this 10-second interval, we can use the formula:
average velocity = (initial velocity + final velocity) / 2
Given that the initial velocity is zero and the final velocity is 98 m/s (as calculated above), we have:
average velocity = (0 + 98 m/s) / 2
average velocity = 49 m/s
So, the average velocity of the object during this 10-second interval is 49 m/s downward.
To determine how far the object will fall during this time, we can use the equation:
distance = (initial velocity * time) + (1/2 * acceleration * time^2)
Since the initial velocity is zero, the equation simplifies to:
distance = 1/2 * acceleration * time^2
Substituting the values, we have:
distance = 1/2 * (9.8 m/s²) * (10 s)^2
distance = 1/2 * 9.8 m/s² * 100 s²
distance = 490 m
Therefore, the object will fall a distance of 490 meters during the 10-second interval.
Regarding your answers, the average velocity of 50 m/s is correct, but the distance of 500 meters is not accurate. It should be 490 meters as calculated above.