Is this equation involving the acceleration of gravity correct?

AI Thread Summary
The discussion centers on the correctness of the equations for gravitational acceleration, g, derived from various kinematic equations. The equations presented assume constant acceleration, specifically in the context of free fall, where initial velocity is zero. It is noted that while some equations may hold true under specific conditions, they are not universally applicable without defining the context and variables involved. The importance of clarifying assumptions, such as vertical motion and the absence of other forces, is emphasized. Overall, the equations can be valid but require proper context to be accurately applied.
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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