Projectiles: Ideal Motion vs. Resisted Motion

Problem Statement

In introductory physics, projectile motion is often modelled purely under the influence of gravity, resulting in a perfect, symmetrical parabolic trajectory. However, real-world objects move through fluid mediums (like air or water) and are subject to resistance. The trajectory of a projectile moving subject to resistance is fundamentally different—it is asymmetrical, with a shorter range, a steeper angle of descent, and a lower maximum height.

Understanding this difference, and proving it mathematically, is a core component of the HSC Mechanics syllabus.

Consider a particle of mass $m$ projected vertically upwards from the ground with an initial velocity $U$.

(a) Assuming no air resistance, derive an expression for the maximum height $H_{ideal}$ reached by the particle.

(b) Now assume the particle experiences air resistance proportional to its velocity, such that the resistive force is $R = mkv$, where $k$ is a positive constant. Derive an expression for the time $T$ taken to reach the maximum height.

(c) For the resisted motion case in (b), show that the maximum height $H_{resisted}$ is given by: $H_{resisted} = \frac{U}{k} - \frac{g}{k^2} \ln \left( 1 + \frac{kU}{g} \right)$

Hints

• Part (a): Use the standard equation of motion $\ddot{y} = -g$. Integrate using the $v \frac{dv}{dy}$ form of acceleration, or use the standard kinematic formulas, setting $v=0$ at max height.
• Part (b): Setup the differential equation for upward motion with resistance: $m\ddot{y} = -mg - mkv$. Use $\frac{dv}{dt}$ for acceleration to find a relationship between velocity and time. Integrate and apply initial conditions to find $t$, then set $v=0$ to find the time to max height.
• Part (c): To find max height, you need the relationship between velocity and displacement. Use the alternate form of acceleration: $v \frac{dv}{dy} = -(g + kv)$. Separate variables, integrate from $y=0$ to $y=H_{resisted}$, and use logarithms to solve.

Solutions

Part (a): Ideal Maximum Height

1. Without air resistance, the only force is gravity downwards.
2. Acceleration is $a = v \frac{dv}{dy} = -g$.
3. Integrate with respect to $y$: $\int v \, dv = \int -g \, dy$ $\frac{1}{2}v^2 = -gy + C$
4. Apply initial conditions: At $y=0$, $v=U$. $\frac{1}{2}U^2 = 0 + C \implies C = \frac{1}{2}U^2$ $\frac{1}{2}v^2 = -gy + \frac{1}{2}U^2$
5. At maximum height $y = H_{ideal}$, the velocity $v = 0$: $0 = -g H_{ideal} + \frac{1}{2}U^2$ $g H_{ideal} = \frac{1}{2}U^2$ $H_{ideal} = \frac{U^2}{2g}$

Part (b): Time to Max Height (Resisted)

1. With air resistance, the net force upwards is $-mg - mkv$.
2. Equation of motion: $m \frac{dv}{dt} = -m(g + kv) \implies \frac{dv}{dt} = -(g + kv)$.
3. Separate variables to find time: $\frac{dt}{dv} = -\frac{1}{g + kv}$ $t = \int -\frac{1}{g + kv} \, dv$
4. Integrate: $t = -\frac{1}{k} \ln (g + kv) + C$
5. Apply initial conditions: At $t=0$, $v=U$. $0 = -\frac{1}{k} \ln (g + kU) + C \implies C = \frac{1}{k} \ln (g + kU)$ $t = \frac{1}{k} \ln (g + kU) - \frac{1}{k} \ln (g + kv) = \frac{1}{k} \ln \left( \frac{g + kU}{g + kv} \right)$
6. At maximum height, $v = 0$. Substitute to find $T$: $T = \frac{1}{k} \ln \left( \frac{g + kU}{g + 0} \right) = \frac{1}{k} \ln \left( 1 + \frac{kU}{g} \right)$

Part (c): Maximum Height (Resisted)

1. We need the relationship between velocity and displacement. Start with the equation of motion and use $a = v \frac{dv}{dy}$: $v \frac{dv}{dy} = -(g + kv)$
2. Separate variables: $\frac{dy}{dv} = -\frac{v}{g + kv}$ $\int dy = \int -\frac{v}{g + kv} \, dv$
3. To integrate the right side, perform an algebraic trick. Add and subtract $g/k$ in the numerator, or manipulate the fraction: $-\frac{v}{g + kv} = -\frac{1}{k} \left( \frac{kv}{g + kv} \right) = -\frac{1}{k} \left( \frac{g + kv - g}{g + kv} \right) = -\frac{1}{k} \left( 1 - \frac{g}{g + kv} \right)$
4. Now integrate from $y=0$ (where $v=U$) to $y=H_{resisted}$ (where $v=0$): $[y]_0^{H_{resisted}} = -\frac{1}{k} \int_U^0 \left( 1 - \frac{g}{g + kv} \right) \, dv$
5. Reverse the limits to absorb the negative sign: $H_{resisted} = \frac{1}{k} \int_0^U \left( 1 - \frac{g}{g + kv} \right) \, dv$
6. Integrate: $H_{resisted} = \frac{1}{k} \left[ v - \frac{g}{k} \ln (g + kv) \right]_0^U$
7. Evaluate at limits: $H_{resisted} = \frac{1}{k} \left( (U - \frac{g}{k} \ln (g + kU)) - (0 - \frac{g}{k} \ln(g)) \right)$ $H_{resisted} = \frac{U}{k} - \frac{g}{k^2} \ln (g + kU) + \frac{g}{k^2} \ln(g)$
8. Combine logarithms: $H_{resisted} = \frac{U}{k} - \frac{g}{k^2} (\ln (g + kU) - \ln(g))$ $H_{resisted} = \frac{U}{k} - \frac{g}{k^2} \ln \left( \frac{g + kU}{g} \right)$ $H_{resisted} = \frac{U}{k} - \frac{g}{k^2} \ln \left( 1 + \frac{kU}{g} \right)$

Takeaways

• Impact of Resistance: Mathematical modeling proves what intuition suggests: air resistance fundamentally alters the kinematics of a projectile, leading to complex logarithmic relationships for time and displacement rather than simple quadratics.
• Algebraic Manipulation of Integrands: Integrating expressions like $\frac{v}{g+kv}$ is a common stumbling block. Learning the trick of rewriting the numerator to mirror the denominator (e.g., adding and subtracting a constant) is essential.
• Connecting Acceleration Forms: Depending on whether you need to link velocity to time or velocity to displacement, you must fluently switch between $a = \frac{dv}{dt}$ and $a = v \frac{dv}{dy}$.

Further Readings

• HSC Mechanics: https://vumaths.com/booklets/hsc-mechanics/
• HSC Integrals: https://vumaths.com/booklets/hsc-integrals/
• HSC Differential Equations: https://vumaths.com/booklets/hsc-differential-equations/

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