- #1

tryingtolearn1

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- Homework Statement
- There are two questions in my book that are almost the same but have two different answers.

The first question:

A particle is projected vertically upward in a constant gravitational field with an initial speed of ##v0##. Show that if there is a retarding force proportional to the square of the speed, the speed of the particle when it returns to its initial position is where $$\frac{v_ov_T}{\sqrt{v_o^2+v_T^2}}$$ ##v_T## is the terminal speed.

Second question:

A baseball is thrown vertically up with speed v o and is subject to a quadratic drag with magnitude ##f (v) = cv^2## . Write down the equation of motion for the upward journey (measuring ##y## vertically up).

- Relevant Equations
- ##F=ma##

But the answers to these two questions confuses me.

For the first question the answer goes like: $$m\ddot y = -mkv^2-mg$$ $$\therefore \ddot y = v\frac{dv}{dy}$$ $$\therefore v\frac{dv}{dy}= -kv^2-g$$

but for the answer to the second question we have: $$m\dot v = mg - cv^2.$$

Both questions are very similar but with different answers. Why is the first question solving it using ##\frac{dv}{dy}\frac{dy}{dt}## instead of just ##\dot v##?

For the first question the answer goes like: $$m\ddot y = -mkv^2-mg$$ $$\therefore \ddot y = v\frac{dv}{dy}$$ $$\therefore v\frac{dv}{dy}= -kv^2-g$$

but for the answer to the second question we have: $$m\dot v = mg - cv^2.$$

Both questions are very similar but with different answers. Why is the first question solving it using ##\frac{dv}{dy}\frac{dy}{dt}## instead of just ##\dot v##?