IB Physics Air Resistance is Small (but not negligible)

Most textbook problems state that air resistance, the drag force on a projectile due to friction as it moves through a fluid medium, is to be neglected. A tutorial sheet of “show that” questions for the case where the drag force is small compared to the weight are given below. All questions give approximate solutions, correct to the first order in k/m, where k is a small positive number called the drag constant and m is the mass of the particle.

1. A particle of mass m falls vertically from rest through air. If the drag force is given by f_d = -k v, where v is the speed of the object, show that the distance fallen at time t is given by s = 1/2 gt²( 1 - kt/(3m) ).
2. A particle of mass m falls vertically from rest through air. If the drag force is given by f_d = -k v², where v is the speed of the object, show that the distance fallen at time t is given by s = 1/2 gt²( 1 - gkt²/(6m) ).
3. A particle of mass m is projected at a speed u at an angle 𝛼 to the horizontal in air where the drag force is proportional to the velocity vector of the particle, f_d = -k v⃗, where k is a small positive constant. Show that the time taken to reach the highest point is given by t = u sin𝛼/g - k u²sin²𝛼/(2mg²).
4. A particle of mass m is projected at a speed u at an angle 𝛼 to the horizontal in air where the drag force is proportional to the velocity vector of the particle, f_d = -k v⃗, where k is a small positive constant. Show that the path of the projectile is y = x tan𝛼 - g x²sec²𝛼/(2u²) + k/m( x²sin𝛼/(2ucos²𝛼) - 2/3x³/(ucos𝛼)³ )

IB Physics Projectile Thrown from a Cliff

The maximum range of a projectile on the horizontal level of its projection (neglecting air resistance) occurs when the angle of projection to the horizontal is 45°. A more difficult problem is when the point of projection is above the horizontal plane on which the object lands. A tutorial sheet of difficult “show that” problems is given below.

1. A projectile is thrown at an angle 𝜽 to the horizontal at a speed u from a vertical height h. Show that the horizontal range of the projectile is u²/(2g)( sin2𝜽 +√(sin²2𝜽 +8ghcos²𝜽/u²) ).
2. A projectile is thrown at a speed u from a height h above horizontal ground. Show that the angle of projection to the horizontal for maximum range on the ground is tan^-1√ (u²/(u²+2gh) ).
3. A projectile is thrown at a speed u from a height h above horizontal ground. Show that the maximum range of the projectile is u/g√(u²+2gh).
4. A projectile is thrown from a vertical height above the ground and maximum range occurs when the angle of projection to the horizontal is 𝜽. Show that the angle (in the maximum range case only) at which the projectile strkes the horizontal is 90° - 𝜽.