Now what about aiming projectiles up and down a hill ??

We have to shift our normal horizontal axis and place it on the hill

So now gravity does not act only in the 'y' or 'j' plane but it also acts parallel to the hill, too.

so we need to resolve the acceleration due to gravity into 2 components, one parallel to the hill and the other perpendicular to the hill

Projectiles on the inclined plane

The key thing here is to let the inclined plane be the x-axis (or the i-axis).

If the projectile is projected at an angle

to the plane, which is itself inclined at an angle to the horizontal, then the four key equations will be:


and (If the particle travels down the plane, the equations will be as above, except with a plus instead of a minus in the equations for vx and Sx )

• To find the range along the plane, find Sx when Sy = 0

• To find the maximum perpendicular height above the plane, find Sy when vy = 0

• If the particle lands perpendicularly to the plane, then vx = 0 when Sy = 0

• The landing angle (l) is given by when Sy = 0

• If the particle lands while travelling in a horizontal direction, then when Sy = 0 (since the landing angle and the angle of the hill will be the same)

Projectiles which bounce

The first step is to find vx and vy for the moment when the particle lands from the first ‘flight’. As the particle takes off on its second flight, the initial velocities will be given by

ux = vx (from first flight) and uy = (-e)vy (from first flight). Now proceed as before.

Common Mistakes

• Lack of basic trigonometry. E.g. how do you find two viable solutions to the equation ? (The answers are 15º and 75º)

• Mixing up

and .

• Lack of familiarity with the formulae on Page 9 of the Mathematical Tables.

Landing Angles are next

Past Papers Questions