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🏔️ Projectile from a Height

📖 The Physics

A projectile has two independent motions at once: a constant horizontal velocity vₓ = v·cosθ (no horizontal force) and a vertical motion controlled only by gravity, starting from v_y = v·sinθ. When you throw from a height h, the ball simply has extra distance to fall, so it stays in the air longer and lands beyond the base.

Taking the ground as y = 0 and up as positive, the position at any time t is:

x=vcos⁡θ  tx = v\cos\theta \; t
y=h+vsin⁡θ  t−12 g t2y = h + v\sin\theta \; t - \tfrac{1}{2}\, g\, t^{2}

Landing time — set y = 0 and solve the quadratic:

12 g t2−vsin⁡θ  t−h=0\tfrac{1}{2}\, g\, t^{2} - v\sin\theta \; t - h = 0
t=vsin⁡θ+(vsin⁡θ)2+2ghgt = \dfrac{v\sin\theta + \sqrt{(v\sin\theta)^{2} + 2gh}}{g}

Then the range and impact speed are:

R=vcos⁡θ⋅tvimpact=vx2+vy2R = v\cos\theta \cdot t \qquad v_{\text{impact}} = \sqrt{v_x^{2} + v_y^{2}}

The five cases are all the same equations — only θ and h change. Horizontal uses θ = 0; from the ground uses h = 0 (giving R = v²·sin2θ/g); throwing down from a height makes θ negative; straight up uses θ = 90° so vₓ = 0 and it drops back down.

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