Is this equation involving the acceleration of gravity correct?

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SUMMARY

The discussion centers on the validity of the equation for gravitational acceleration, represented as $$g=\dfrac{v}{t}=\dfrac{2s}{t^{2}}=\dfrac{2\left( vt-s\right) }{t^{2}}=\dfrac{v^{2}}{2s}=9.8\ \mathrm{m s^{-2}}$$. Each equation is derived from the equations of motion under uniform acceleration, where variables are defined as follows: s = displacement, u = initial velocity, v = final velocity, a = acceleration, and t = time. The consensus is that while these equations can be valid under specific conditions, they are not universally applicable without additional context regarding the motion's parameters.

PREREQUISITES
  • Understanding of kinematic equations for uniformly accelerated motion
  • Familiarity with the concepts of displacement, initial velocity, final velocity, and acceleration
  • Basic knowledge of gravitational acceleration and its standard value (9.8 m/s²)
  • Ability to analyze motion in a vertical direction under the influence of gravity
NEXT STEPS
  • Study the derivation and application of the equations of motion in physics
  • Learn how to apply the concept of uniform acceleration in real-world scenarios
  • Explore the implications of initial conditions on the outcomes of kinematic equations
  • Investigate the role of forces acting on objects in vertical motion, particularly gravity
USEFUL FOR

Students of physics, educators teaching kinematics, and anyone interested in understanding the principles of motion under gravity will benefit from this discussion.

Huzaifa
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Homework Statement
Is this equation correct: $$g=\dfrac{v}{t}=\dfrac{2s}{t^{2}}=\dfrac{2\left( vt-s\right) }{t^{2}}=\dfrac{v^{2}}{2s}=9.8ms^{-2}$$?
Relevant Equations
##g=\dfrac{v}{t}=\dfrac{2s}{t^{2}}=\dfrac{2\left( vt-s\right) }{t^{2}}=\dfrac{v^{2}}{2s}=9.8ms^{-2}##
$$\begin{aligned}v=u+at\\ \Rightarrow v=gt\\ \Rightarrow g=\dfrac{v}{t} \cdots (1)\end{aligned}$$
$$\begin{aligned}s=ut+\dfrac{1}{2}at^{2}\\ \Rightarrow s=\dfrac{1}{2}gt^{2}\\ \Rightarrow g=\dfrac{2s}{t^{2}}\cdots (2)\end{aligned}$$
$$\begin{aligned}s=vt-\dfrac{1}{2}at^{2}\\ \Rightarrow s-vt=-\dfrac{1}{2}at^{2}\\ \Rightarrow g=\dfrac{2\left( vt-s\right) }{t^{2}} \cdots (3)\end{aligned}$$
$$\begin{aligned}v^{2}-u^{2}=2as\\ \Rightarrow v^{2}=2as\\ \Rightarrow g=\dfrac{v^{2}}{2s} \cdots (4)\end{aligned}$$
$$g=\dfrac{v}{t}=\dfrac{2s}{t^{2}}=\dfrac{2\left( vt-s\right) }{t^{2}}=\dfrac{v^{2}}{2s}=9.8\ \mathrm{m s^{-2}}$$
 
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What's the question precisely?
 
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Huzaifa said:
Homework Statement:: Is this equation correct: $$g=\dfrac{v}{t}=\dfrac{2s}{t^{2}}=\dfrac{2\left( vt-s\right) }{t^{2}}=\dfrac{v^{2}}{2s}=9.8ms^{-2}$$?
Relevant Equations:: ##g=\dfrac{v}{t}=\dfrac{2s}{t^{2}}=\dfrac{2\left( vt-s\right) }{t^{2}}=\dfrac{v^{2}}{2s}=9.8ms^{-2}##

Huzaifa said:
$$\begin{aligned}v=u+at\\ \Rightarrow v=gt\\ \Rightarrow g=\dfrac{v}{t} \cdots (1)\end{aligned}$$
$$\begin{aligned}s=ut+\dfrac{1}{2}at^{2}\\ \Rightarrow s=\dfrac{1}{2}gt^{2}\\ \Rightarrow g=\dfrac{2s}{t^{2}}\cdots (2)\end{aligned}$$
$$\begin{aligned}s=vt-\dfrac{1}{2}at^{2}\\ \Rightarrow s-vt=-\dfrac{1}{2}at^{2}\\ \Rightarrow g=\dfrac{2\left( vt-s\right) }{t^{2}} \cdots (3)\end{aligned}$$
$$\begin{aligned}v^{2}-u^{2}=2as\\ \Rightarrow v^{2}=2as\\ \Rightarrow g=\dfrac{v^{2}}{2s} \cdots (4)\end{aligned}$$
$$g=\dfrac{v}{t}=\dfrac{2s}{t^{2}}=\dfrac{2\left( vt-s\right) }{t^{2}}=\dfrac{v^{2}}{2s}=9.8\ \mathrm{m s^{-2}}$$
What you have is multiple equations.

You need to describe the overall situation. Also define precisely what each variable means.

Of course, answer the question asked by @PeroK. .
 
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SammyS said:
You need to describe the overall situation. Also define precisely what each variable means.
The overall situation is constant or uniform acceleration. Here, s = displacement, u = initial velocity, v = final velocity, a = acceleration, t = time. These are equations of motion.
 
None of the numbered equations [except maybe (3)] are generally true.
Each is true in some special [so far unstated] cases
 
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Hi @Huzaifa. You seem to be making a number of unstated assunptions, e.g.
- motion is in the vertical direction only, with gravity the only force;
- initial velocity (u) is zero, so you are only considering objects released from rest at t=0.

But of course I'm just guessing, as you haven't yet answered @PeroK's question (Post #2).
 
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