Prove in case of projectile d^2 (v^2 ) / dt^2 = 2g^2

AI Thread Summary
The discussion focuses on proving the equation d^2(v^2)/dt^2 = 2g^2 for a projectile. Participants emphasize the need for clarity in notation and proper application of calculus, specifically the Chain Rule, to derive the second derivative of v^2. A common misunderstanding is highlighted, where the derivative of v^2 is incorrectly stated as "2v" instead of the correct form, which includes the term dv/dt. The conversation encourages detailed explanations of each step in the solution process. Overall, the importance of using standard notation and accurate mathematical principles in physics problems is underscored.
Safi Majid
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Homework Statement


for a projectile show that d^2 (v^2) / dt^2 = 2g^2

2. The attempt at a solution
=d/dt (d(v^2)/dt)
=d/dt (2v)
 
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Your attempt is using a very strange notation. Please write out the details using standard notation and explain what you are doing in each step.
 
Safi Majid said:

Homework Statement


for a projectile show that d^2 (v^2) / dt^2 = 2g^2

2. The attempt at a solution
=d/dt (d(v^2)/dt)
=d/dt (2v)

Welcome to PF!
The velocity v depends on t. You need to apply the Chain Rule to get the time derivative of v2.
 
Orodruin said:
Your attempt is using a very strange notation. Please write out the details using standard notation and explain what you are doing in each step.
i have used second derivative on it
 
In standard notation it is
d2(v2)/dt2=2g2
 
That was not what was being questioned. The derivative of v^2, with respect to t, is NOT "2v". It is 2v\frac{dv}{dt}.
 
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