what is the instantaneous velocity of a free falling object 10s after it is released from a postion of rest?
what is its average velocity during this 10s interval?
how far will it fall during this time?
To find the instantaneous velocity of a free-falling object 10 seconds after it is released from a position of rest, you can use the equation for the velocity of a falling object:
v = gt
where v is the velocity, g is the acceleration due to gravity (approximately 9.8 m/s²), and t is the time.
Plugging in the values, we get:
v = (9.8 m/s²)(10 s) = 98 m/s
Therefore, the instantaneous velocity of the object 10 seconds after being released is 98 m/s.
To find the average velocity during the 10-second interval, you need to know the initial velocity and the final velocity. In this case, the initial velocity is zero (since the object starts from rest), and the final velocity is 98 m/s (as calculated above).
The average velocity is given by the formula:
average velocity = (initial velocity + final velocity) / 2
Plugging in the values:
average velocity = (0 m/s + 98 m/s) / 2 = 49 m/s
Therefore, the average velocity during the 10-second interval is 49 m/s.
To calculate the distance the object will fall during this time, you can use the equation for the distance traveled by a falling object:
d = (1/2)gt²
where d is the distance, g is the acceleration due to gravity, and t is the time.
Plugging in the values:
d = (1/2)(9.8 m/s²)(10 s)² = (1/2)(9.8 m/s²)(100 s²) = 490 m
Therefore, the object will fall a distance of 490 meters during this 10-second interval.
for constant acceleration:
V = Vo + a t
so for a = g = -9.8 m/s^2
V = 0 - 9.8 t = -98 meters/second
average = (Vo+V)/2 = (0-98)/2 = 49 m/s
distance = average velocity * time = 490 meters