# Differential equation: projectiles

In summary, the conversation discusses the derivation of an equation of motion for a vertically projected projectile with a resistive force due to air resistance. A differential equation for velocity as a function of height is derived and solved to find the maximum height reached by the projectile. The equation differs from traditional projectile motion due to the inclusion of the resistive force.

## Homework Statement

A projectile of mass m is projected vertically upwards with speed U. In addition to its
weight it experiences a resistive force mkv^2, where v is the speed at which the projectile is moving and k is a constant.
Derive the equation of motion of the projectile:

dv/dt = -(g + kv^2)

Derive a differential equation for v as a function of z, where z is the height of the projectile above its starting position. Solve this equation to find the maximum height reached by the projectile.

## The Attempt at a Solution

F = ma = m dv/dt

-mg - mkv^2 = m dv/dt

dv/dt = -g - kv^2 = -(g + kv^2)

dv/dt = dv/dz * dz/dt = v dz/dt

v dv/dz = -(g + kv^2)

[v/(g+kv^2)] dv = - dz

Integrating:

1/2k ln(g + kv^2) = -z + C

ln(g + kv^2) = -2kz + 2kC

When v = U, z = 0, 2kC = ln(g + kU^2)

ln(g + kv^2) = -2kz + ln(g + kU^2)

2kz = ln([g+kU^2]/[g+kv^2])

Maximum z is when v = 0.
z = 1/2k ln((g + kU^2)/(g)) = 1/2k ln(1 + kU^2/g)

Is this correct?

Thankyou.

Seems OK. Note that this is the eqn when the projectile is going up. When it's falling back, the eqn would be slightly different, because of the v^2 dependence of acceleration, unlike projectile motion without air resistance.

Yes, this appears to be a correct derivation. To solve for the maximum height, you can set the derivative of z with respect to v equal to 0 and solve for v. This will give you the value of v at the maximum height, which can then be plugged into the equation for z to find the maximum height.

## 1. What is a differential equation?

A differential equation is a mathematical equation that relates one or more unknown functions to their derivatives. It is used to model and analyze various physical phenomena, such as projectiles in motion.

## 2. How are differential equations used to model projectiles?

Differential equations are used to model projectiles by describing the relationship between the projectile's position, velocity, and acceleration over time. This allows us to predict the trajectory and other characteristics of the projectile's motion.

## 3. What is the difference between a first-order and a second-order differential equation?

A first-order differential equation involves only the first derivative of the unknown function, while a second-order differential equation involves the second derivative. In the context of projectiles, a first-order differential equation is used to model the projectile's velocity, while a second-order equation is used to model its position.

## 4. How do initial conditions affect the solution of a projectile's differential equation?

Initial conditions, such as the initial position and velocity of the projectile, are used to determine the specific solution to the projectile's differential equation. This means that even a small change in initial conditions can result in a significantly different trajectory for the projectile.

## 5. What is the significance of the "drag" term in a projectile's differential equation?

The "drag" term in a projectile's differential equation represents the force of air resistance acting on the projectile. This term is important in accurately modeling the projectile's motion, as it can significantly affect its trajectory and speed.

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