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A ball is thrown from the foot of a hill and lands somewhere up the slope. You know projectile motion on flat ground, but here the usual range formula gives the wrong answer and the algebra with horizontal axes gets messy fast. This post shows how tilting your axes turns the problem into two short calculations, with one full example and the mistakes examiners expect.
Why do the usual projectile formulas fail on a slope?
The formulas you memorised, such as R = u² sin 2θ / g, assume the ball lands at the same height it started from. On a slope, the landing point is higher (or lower) than the launch point, and where it lands depends on the shape of the hill. The formulas are not wrong; they answer a different question.
What does carry over is the method. NCERT's Class 11 chapter on motion in a plane treats a projectile as two independent motions, one along each axis, each with constant acceleration. OpenStax University Physics makes a useful point: horizontal and vertical axes are the most sensible choice on flat ground, but "it is not required that we use this choice of axes; it is simply convenient." On a slope, a different choice is more convenient.
How do you set up axes along the incline?
Let the slope make angle α with the horizontal, and let the ball leave at speed u, at angle θ measured from the slope (not from the ground).
x-axis along the slope, pointing up it. Initial velocity u cos θ. Acceleration −g sin α.
y-axis perpendicular to the slope, pointing away from it. Initial velocity u sin θ. Acceleration −g cos α.
Now the ball lands when y = 0 again, exactly like flat-ground motion. Both motions have constant acceleration, so the equations of motion apply along each axis separately.
From y = 0 at landing, the time of flight is T = 2u sin θ / (g cos α). The range along the slope is R = (u cos θ)T − ½(g sin α)T².
Worked example: a throw up the slope
A ball is thrown from the foot of a 30° slope at 20 m/s, at 30° above the slope. Take g = 10 m/s². Find the time of flight, the distance up the slope where it lands, and the greatest distance from the slope.
Components of g: along the slope g sin 30° = 5 m/s²; perpendicular g cos 30° ≈ 8.66 m/s².
Components of u: along the slope 20 cos 30° ≈ 17.32 m/s; perpendicular 20 sin 30° = 10 m/s.
Time of flight: T = 2 × 10 / 8.66 ≈ 2.31 s.
Range along the slope: R = 17.32 × 2.31 − ½ × 5 × 2.31² ≈ 40.0 − 13.3 ≈ 26.7 m.
Greatest distance from the slope, when the perpendicular velocity is zero: H = 10² / (2 × 8.66) ≈ 5.77 m.
Check with ordinary axes. The launch is 60° above the horizontal, so horizontal speed 10 m/s and vertical speed 17.32 m/s. The ball is on the slope when its height equals x tan 30°: 17.32t − 5t² = 10t × 0.577. That gives t ≈ 2.31 s and x ≈ 23.1 m, and 23.1 / cos 30° ≈ 26.7 m along the slope. Same answer, more algebra.
What angle gives the maximum range on an incline?
Working through the range formula with a little trigonometry gives R = 2u² sin θ cos(θ + α) / (g cos² α). This is largest when θ = 45° − α/2, and the maximum range up the slope is u² / [g(1 + sin α)].
For our slope, 45° − 15° = 30°, which is the angle we used. The maximum range is 400 / (10 × 1.5) ≈ 26.7 m, matching step 4. For a throw down the slope, replace α with −α: the best angle is 45° + α/2 and the maximum range is u² / [g(1 − sin α)].
A second classic: when does the ball hit the slope at right angles? Its velocity along the slope must be zero at landing, which works out to cot θ = 2 tan α. On a 30° slope, that is θ ≈ 40.9° above the slope.
What mistakes cost marks in incline projectile questions?
Using the flat-ground range formula. It ignores the slope entirely.
Forgetting that gravity has a component along the slope. With tilted axes, the x-motion is no longer uniform.
Mixing up the two angles. If the question says "at 60° to the horizontal" on a 30° slope, the angle to the slope is 30°.
Wrong signs for a throw down the slope. Going down, g sin α speeds the ball up along x, so its sign flips.
Taking cos α for sin α. A quick test: when α = 0 your answers must become the flat-ground formulas.
If resolving vectors is where the errors start, fix that first; every step here depends on it.
A practice routine for incline projectiles
Draw the slope, the launch velocity and both angles for every problem, even easy ones.
Solve two problems twice: once with tilted axes, once with horizontal axes. Confirm they agree.
Derive the time-of-flight and range formulas yourself once, then check them with α = 0.
Do three throws down the slope in a row, since the sign change is where most errors hide.
Mix in two flat-ground projectiles so you practise choosing the axes, not just using them.
If you want to see how the path and the landing point move as you change the angle, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read. Our projectile simulation activity gives you predictions to test.
Today, take one incline problem you got wrong, redraw it with tilted axes and solve it again.
