Maximum velocity of object dropped from cliff

  • Thread starter Thread starter Jigen
  • Start date Start date
  • Tags Tags
    Maximum Velocity
AI Thread Summary
The maximum velocity of an object dropped from a 20 m cliff, assuming no air friction and gravity of 10 m/s², is 20 m/s. Initial calculations suggested 15 m/s, but this was incorrect due to misunderstanding the average speed during free fall. The object accelerates, reaching 10 m/s after 1 second and 20 m/s after 2 seconds. Using the SUVAT equation V² = U² + 2aS confirms that the final velocity is indeed 20 m/s. Therefore, the correct answer is 20 m/s.
Jigen
Messages
2
Reaction score
0
Just finished an exam. One problem is bothering me and I want to be certain if it's right.

Homework Statement


What is the maximum velocity of an object dropped from a 20 m cliff?
Assume no air friction and gravity is 10 m/s/s.


Homework Equations


None given, but I assume v=gt is relevant.


The Attempt at a Solution


15 m/s.
My reasoning was that if after 1s, the object fell 10m at 10m/s,
and after 2s, the object has fallen 30m (10m + 20m) at 20m/s (10m/s + 10m/s),
then I can divide the 2nd speed increment in half to determine speed at 20m fall (10m/s + 5m/s).

I tried using equations before answering 15.
v=gt by itself can't work without t (t was not given).
d=1/2gt^2 tells me the object fell for 2s, but that doesn't seem right (it would have traveled 30m).
Assuming t=2s is right, v = 10m/s/s * 2s = 20m/s.

So which is right: 15 or 20?
 
Physics news on Phys.org
Jigen said:
Just finished an exam. One problem is bothering me and I want to be certain if it's right.

Homework Statement


What is the maximum velocity of an object dropped from a 20 m cliff?
Assume no air friction and gravity is 10 m/s/s.

Homework Equations


None given, but I assume v=gt is relevant.

The Attempt at a Solution


15 m/s.
My reasoning was that if after 1s, the object fell 10m at 10m/s,
and after 2s, the object has fallen 30m (10m + 20m) at 20m/s (10m/s + 10m/s),
then I can divide the 2nd speed increment in half to determine speed at 20m fall (10m/s + 5m/s).

I tried using equations before answering 15.
v=gt by itself can't work without t (t was not given).
d=1/2gt^2 tells me the object fell for 2s, but that doesn't seem right (it would have traveled 30m).
Assuming t=2s is right, v = 10m/s/s * 2s = 20m/s.

So which is right: 15 or 20?

It's 20.
In the first second it reaches 10m/s, but travels only 5m [average speed = (0+10)/2 = 5
In the second second second is increases to 20 m/s, traveling a further 15m [average speed = (10+20)/2 = 15.
 
Oh, that's right... it has to be averaged since speed isn't constant. Thanks.
 
In my day we would remember one of the the SUVAT Equations

http://en.wikipedia.org/wiki/Equations_of_motion

probably V2 = U2 + 2aS

where

V = final velocity
U = initial velocity
a = g ≈10
s = displacement

Reduces to

V2 = 2as
= 2 * 10 * 20
= 400

so V=20
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Back
Top