---
title: "An object of mass \\(m\\) is launched vertically upward from ground level with an initial speed \\(v_0\\). As the object ascends, it experiences a downward drag force of magnitude \\(F_d = bv\\), where \\(b\\) is a positive constant and \\(v\\) is the speed of the object. Acceleration due to gravity is \\(g\\). Which of the following expressions gives the maximum height \\(y_{\\text{max}}\\)"
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url: "https://nerd-notes.com/ubq/117669/"
date_modified: "2026-08-04T07:49:46+00:00"
---

# An object of mass \(m\) is launched vertically upward from ground level with an initial speed \(v_0\). As the object ascends, it experiences a downward drag force of magnitude \(F_d = bv\), where \(b\) is a positive constant and \(v\) is the speed of the object. Acceleration due to gravity is \(g\). Which of the following expressions gives the maximum height \(y_{\text{max}}\)

An object of mass \(m\) is launched vertically upward from ground level with an initial speed \(v_0\). As the object ascends, it experiences a downward drag force of magnitude \(F_d = bv\), where \(b\) is a positive constant and \(v\) is the speed of the object. Acceleration due to gravity is \(g\). Which of the following expressions gives the maximum height \(y_{\text{max}}\)

![A small sphere labeled m moving vertically upward with an upward velocity vector labeled v_0. Two downward force vectors originate from the center of the sphere: one labeled mg pointing straight down, and another labeled F_d = bv pointing straight down. A vertical coordinate axis points upward with y = 0 at the launch level and y_max at the peak. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-drag-setup-1785829786-bdLI8X.jpg)

- **A.** \(\dfrac{m v_0}{b} - \dfrac{m^2 g}{b^2} \ln\left(1 + \dfrac{b v_0}{m g}\right)\)
- **B.** \(\dfrac{m v_0}{b} + \dfrac{m^2 g}{b^2} \ln\left(1 - \dfrac{b v_0}{m g}\right)\)
- **C.** \(\dfrac{m v_0}{b} + \dfrac{m^2 g}{b^2} \ln\left(1 + \dfrac{b v_0}{m g}\right)\)
- **D.** \(\dfrac{m^2 g}{b^2} \ln\left(1 + \dfrac{b v_0}{m g}\right)\)

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/117669/*