Question

A body of constant mass m is projected vertically upward with an initial velocity v0 in a medium offering a resistance k | v | , where k is a constant. Neglect changes in the gravitational force.
i) Find the maximum height xm attained by the body and the time tm at which this maximum height is reached.
ii) Show that if kv0 /mg < 1, then tm and xm can be expressed as

t_{m}= \frac{V_{0}}{g}\left [ 1-\frac{1}{2}\frac{kv_{0}}{mg} + \frac{1}{3}(\frac{kv_{0}}{mg})^2..... \right ]

x_{m}= \frac{V^2_{0}}{2g}\left [ 1-\frac{2}{3}\frac{kv_{0}}{mg} + \frac{1}{2}(\frac{kv_{0}}{mg})^2..... \right ]

15 Feb 2018
Answer :
Word Count : 657
Let's solve the problem step by step. ### i) Maximum Height and Time to Reach Maximum Height: We are given a body of constant mass \( m \) projected vertically upward with an initial velocity \( v_0 \) in a medium that offers a resistance proportional to the velocity, \( k |v| \), where \( k \) is a constant. We neglect changes in the gravitational force. The forces acting on the body are: - Gravitational force: \( F_g = -mg \) - Resistive force due to the medium: \( F_r = -k |v| \) Thus, the total force acting on the body is: \[ F = -mg - k |v| \] Using Newton's second law, \( F = ma = m \frac{dv}{dt} \), we get the equation of motion: \[ m \frac{dv}{dt} = -mg - k |v| \] Dividing both sides by \( m \), we have: \[ \frac{dv}{dt} = -g - \frac{k}{m} |v| \] For upward motion, the velocity is positive, so \( |v| = v \), and the equation becomes: \[ \frac{dv}{dt} = -g - \frac{k}{m} v \] This is a first-order linear differential equation. To solve it, we can separate variables: \[ \frac{dv}{-g - \frac{k}{m} v} = dt \] Now, integrate both sides: \[ \int \frac{dv}{-g - \frac{k}{m} v} = \int dt \] On the left-hand side, we can factor out \( -g \) from the denominator to simplify: \[ \int \frac{dv}{-g \left( 1 + \frac{k}{mg} v \right)} = \int dt \] The integral of the left side is: \[ -\frac{m}{k} \ln \left( 1 + \frac{k}{mg} v \right) \]
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