# Calc-based physics question

• dante714
In summary, the value of g, the gravitational acceleration near Earth, can be expressed in terms of s, the distance between two light beams, and the times of passage of a ball projected through the beams. Using Newton's second law and assuming an evacuated tube, the equation for g is 2s/[(t1-t0)*(t1-t3)].
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.

[/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]

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)]

## 1. What is "Calc-based physics"?

"Calc-based physics" refers to the use of calculus in the study of physics. It involves using mathematical techniques such as derivatives and integrals to analyze and solve problems in the field of physics.

## 2. Why is calculus important in physics?

Calculus is important in physics because it allows us to describe and understand the behavior of physical systems using mathematical models. It provides a powerful tool for analyzing complex phenomena and making predictions about the natural world.

## 3. What topics are typically covered in calc-based physics courses?

Calc-based physics courses typically cover topics such as kinematics, Newton's laws of motion, work and energy, rotational motion, gravitation, oscillations and waves, electric and magnetic fields, and thermodynamics.

## 4. How does calc-based physics differ from algebra-based physics?

The main difference between calc-based and algebra-based physics is the level of mathematical rigor. Calc-based physics uses calculus to analyze and solve problems, while algebra-based physics uses simpler mathematical techniques such as algebra and trigonometry. Calc-based physics is typically more advanced and requires a stronger foundation in mathematics.

## 5. What are some real-world applications of calc-based physics?

Calc-based physics has numerous real-world applications in fields such as engineering, astronomy, and medicine. It is used in the design of structures, vehicles, and machines, in the understanding of celestial bodies and their motion, and in medical imaging techniques such as MRI and CT scans.

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