Differential equation: projectiles

[mattgad]

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.

 

[Shooting Star]

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.

 

[Moetasim]

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.