Multidimensional motion problem studio

Start every problem the same way. Fix an origin and axes, write a position vector r⃗(t) = x(t) î + y(t) ĵ for each moving thing, and leave the unknowns as symbols. Then translate each condition into equations. Passes through the target means x = D and y = H at the same t. Meets the other ball means r⃗_A = r⃗_B, both components, at the same t. A vector condition in the plane gives two scalar equations, but t is itself unknown and uses one of them up. So a target point fixes only one launch unknown: for a mark 30 m away and 6 m up, every speed above a threshold has its own angles. A second condition pins the launch. Demand arrival at t = 2.00 s and the components separate cleanly: vₓ = 30/2.00 = 15.0 m s⁻¹ and vᵧ₀ = (6.0 + 4.905 × 2.00²)/2.00 = 12.8 m s⁻¹, a launch of 19.7 m s⁻¹ at 40.5° above the horizontal.

Eliminating t gives the path y = x tan θ − gx²/(2v₀² cos²θ). Usually θ and v₀ go in and heights come out. Here the point (D, H) is given, so the same equation becomes a condition on the launch. With the angle fixed it gives the speed, v₀² = gD²/(2cos²θ (D tan θ − H)), which needs D tan θ > H: an aim on or below the sightline fails at every speed. A 35° launcher reaching (30 m, 6 m) needs 20.9 m s⁻¹. With the speed fixed, write 1/cos²θ = 1 + tan²θ and the condition is a quadratic in tan θ. At 20 m s⁻¹ it reads 11.04 tan²θ − 30 tan θ + 17.04 = 0, with roots at 38.9° and 62.4° that arrive after 1.93 s and 3.24 s. Lower the speed and the discriminant shrinks until, at vₘᵢₙ = √(g(H + √(D² + H²))) = 18.9 m s⁻¹, the two roots merge at 50.7°. Below that speed no angle reaches the mark.

Two objects meet when r⃗_B(t) − r⃗_A(t) reaches zero. If they share an acceleration, as any two bodies in free flight do, the ½gt² terms cancel and the separation is Δr⃗₀ + Δv⃗ t: a straight line traced at a steady rate. They collide only if Δv⃗ points straight back along Δr⃗₀, and only if that moment comes before either lands. Otherwise find the nearest approach by setting the derivative of |Δr⃗₀ + Δv⃗ t|² to zero, which gives t* = −(Δr⃗₀ ⋅ Δv⃗)/|Δv⃗|². Throw stone A from the origin at (8.0, 14.0) m s⁻¹ and, at the same instant, stone B from a ledge at (20.0, 12.0) m with (−4.0, 5.0) m s⁻¹. Then Δr⃗₀ = (20.0, 12.0) m and Δv⃗ = (−12.0, −9.0) m s⁻¹, so t* = 348/225 = 1.55 s and the stones pass 2.40 m apart, A 9.9 m up and B 8.0 m up. The value of g never touched the miss distance; it only placed the near miss in the air.

Real paths come in pieces: a straight run, a circular arc, a free flight. Solve each piece with its own model and pass position and velocity across each join. The acceleration may jump — a straight has no radial part, an arc has v²/R inward, and once the surface is gone only g is left — while the velocity may not. Two consequences do most of the work. A launch angle is the arc's tangent angle where it ends: a ramp of radius 1.20 m that stops 30° past its lowest point launches at 30°, from 1.20 × (1 − cos 30°) = 0.161 m above that point. And a surface can push but not pull. Over a crest the rider stays on only while a free path would bend at least as tightly as the surface; a free path bends with radius v²/(g cos β), so contact is lost where v² > gR cos β. On a crest of radius 6.0 m the threshold is √(9.81 × 6.0) = 7.7 m s⁻¹, and at 8.5 m s⁻¹ the free path's radius is 7.4 m: the rider lifts off at the top.

A formula from a studio problem can be tested before its number is used. Dimensions: g(H + √(D² + H²)) is L T⁻² × L = L² T⁻², a speed squared. Limits: with H = 0 the minimum speed is √(gD) at 45°, the familiar level-ground best range, 17.2 m s⁻¹ for D = 30 m; with D = 0 it is √(2gH), straight up, 10.8 m s⁻¹ for H = 6 m; and as v₀ grows the low angle root tends to tan θ = H/D, straight along the sightline, as it must once the drop is negligible. Roots: solving y(t) = H gives a rising and a falling arrival, and a negative root lies before the launch. Signs: a basket or a bucket needs vᵧ < 0 on arrival, and both 20 m s⁻¹ launches at the 30 m, 6 m mark pass, arriving at −6.35 m s⁻¹ and −14.0 m s⁻¹. An answer that fails any of these is wrong before the arithmetic is even examined.

A launch plan is only as good as its tolerance to what you did not control. Hold everything else fixed and differentiate the result with respect to one input. For the height on arrival at x = D, dy/dθ = D sec²θ (1 − gD tan θ/v₀²). At 20 m s⁻¹ it is +0.35 m per degree for the 38.9° shot and −0.99 m per degree for the 62.4° lob, and it is zero at the minimum-speed aim, which makes that launch the forgiving one. Speed errors rank the same way: dy/dv₀ = gD²/(v₀³ cos²θ) is 1.8 m per m s⁻¹ low and 5.1 m per m s⁻¹ high. Test the assumptions by comparing what was left out with what was kept. The drag deceleration ½ρC_dAv²/m (air density ρ, drag coefficient C_d, frontal area A) is about 8 m s⁻² for a 57 g tennis ball at 20 m s⁻¹, nearly g, so the model fails; for a 7.26 kg shot at 13 m s⁻¹ it is under 1% of g, and the model holds.