Maximum velocity of object dropped from cliff

[Jigen]

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?

 

[PeterO]

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.

 

[Jigen]

Oh, that's right... it has to be averaged since speed isn't constant. Thanks.

 

[CWatters]

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

 

[Nickie]

I would suggest that you use the correct formula to solve this problem. In this case, the correct formula is v=√(2gh), where v is the maximum velocity, g is the acceleration due to gravity (10 m/s/s in this case), and h is the height of the cliff (20 m). Using this formula, we can calculate the maximum velocity to be approximately 19.8 m/s. Therefore, the answer is neither 15 nor 20, but rather closer to 20. I would also recommend double-checking your calculations and making sure that you are using the correct units (meters and seconds) in your equations. Hope this helps!