How Can Gravitational Acceleration 'g' Be Expressed Using Ball Passage Times?

Click For Summary
The discussion focuses on determining the gravitational acceleration 'g' using the passage times of a ball in an evacuated tube. The problem involves measuring the ball's upward and downward flight times through two light beams spaced a known distance 's' apart. The user attempts to relate these times to the motion of the ball under gravity, utilizing Newton's second law. The derived formula for 'g' is presented as g = 2s / [(t1-t0)(t1-t3)], which expresses gravitational acceleration in terms of the distance and time intervals. Confirmation of the solution's accuracy is requested from other participants.
dante714

Homework Statement



In a physics laboratory the value of g , the gravitational acceleration in the vicinity of Earth, has been measured accurately by projecting a ball up an evacuated tube and electronically timing the passage of the ball in its upward and downward flight through two light beams, an accurately known distance 's' apart. [/B]

If the successive times of passage through the beams are: t0 ,t1 ,t2 ,and t3, express 'g' in terms of 's' and the times of the passage of the ball.


Homework Equations


[/B]
Fnet = ma

The Attempt at a Solution



I believe that Newton's 2nd law: Fnet = ma is imperative to solving this problem but I am not exactly sure how. Can anyone answer this problem?[/B]
 
Physics news on Phys.org
You will need to provide a little more of an attempted solution than that. Ask yourself how the times should relate to the motion of the ball if the ball is moving only under the influence of gravity in a gravitational field ##g##.
 
Orodruin said:
You will need to provide a little more of an attempted solution than that. Ask yourself how the times should relate to the motion of the ball if the ball is moving only under the influence of gravity in a gravitational field ##g##.

Hello Orodruin,

This is what I have done so far

Setting u0 as an unknown at t0, s0
u1 at t1, s
u2 at t2, smax-s
u3 at t3, smax

As the tube is evacuated it should be expected that u3 = -u0 and u2 = -u1 - a quick check of t3-t2 and t1-t0 should confirm whether this is a good assumption (it will be closer than a similar experiment in air).

-u0 = u0 + g (t3-t0) :- u0 = 0.5 g (t0-t3)
{-u1 = u1 + g (t2-t1), and u1 = u0 + g (t1-t0) :- u0 + g (t1-t0) = 0.5 g (t2-t1) * not used in this solution}

s = u0 (t1-t0) + 0.5 g (t1-t0)^2
s = [0.5 g (t0-t3)](t1-t0)+0.5 g (t1-t0)^2
g [(t0-t3)+(t1-t0 )] = 2 s / (t1-t0)
g * (t1-t3) = 2s / (t1-t0)

g = 2 s /[(t1-t0)*(t1-t3)]

Please PM me to confirm if this is the correct answer
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
4
Views
3K
  • · Replies 42 ·
2
Replies
42
Views
6K
Replies
39
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
3
Views
3K
  • · Replies 17 ·
Replies
17
Views
3K
Replies
2
Views
2K