What is the instantaneous velocity of a freely falling object 10s after it is released from rest? What is its average velocity during this 10s interval? How far will it fall during this time?
For the average speed I got 50m/s nd for the last one I got 500m
is this right? how do I do the first one? please help
If you use the correct value for g, 9.8 m/s^2, you will find that the instantaneous velocity after 10 seconds is 98 m/s. The average velocity is half that.
The distance travelled is time x (avg. velocity) = 490 m
You method of soloution is correct, you just used a different value of g that you may have been told to use.
Thanks for showing your work. I wish more students would do that.
thank you
my teacher just told me to round it up to 10m/s^2
What is the instantaneous velocity of a falling object 5s after it is released from rest? What it its average velocity during this 5s–interval? How far will it fall during this time?
Given:
T=10s
G= 10m/s
V=?
V=(10 • 10)
V= 100 m/s
Average Velocity= 100/2
So, = 50m/s
Distance= 1/2(10)(10)=50 m
Yay<3
Oh boy, it sounds like you need a little help with physics! Don't worry, Clown Bot is here to add some humor and assist you.
The instantaneous velocity of a freely falling object can be calculated using the equation v = gt, where g is the acceleration due to gravity (9.8 m/s^2) and t is the time. So, after 10 seconds, the instantaneous velocity would be v = 9.8 m/s^2 × 10 s = 98 m/s.
As for the average velocity during the 10-second interval, we can consider it as the displacement divided by the time. Since the object is freely falling, its average velocity would be 0 m/s because it starts at rest and ends at rest.
Now, let's determine how far it will fall during this time. We can use the equation s = (1/2)gt^2, where s is the distance fallen. Plugging in the values, we have s = (1/2) × 9.8 m/s^2 × (10 s)^2 = 490 m.
So, to recap:
- The instantaneous velocity after 10 seconds is 98 m/s.
- The average velocity during the 10-second interval is 0 m/s.
- The distance fallen during this time is 490 m.
Hope that helps! Remember, physics can be fun when you add a touch of clowning around.
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.