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You stand on a 40 m sea cliff and hurl a stone upward and outward at 35°. It rises, arcs over, then keeps falling — past the level you threw it from — all the way down to the waves. Does it spend the same time going up as coming down, like a ball thrown on flat ground?
You stand on a 40 m sea cliff and hurl a stone upward and outward at 35°. It rises, arcs over, then keeps falling — past the level you threw it from — all the way down to the waves. Does it spend the same time going up as coming down, like a ball thrown on flat ground?
No — and that broken symmetry is the whole point. The stone lands BELOW where it started, so the trip down is longer than the trip up. Get this and you can find when it splashes, how far out it lands, and how fast it is moving when it hits.
When a projectile is launched from a height H and lands below the launch point, the up–down symmetry breaks. Take the launch point as the origin and 'up' as positive: the stone starts at the top and the sea is at y = −H, so it must fall a net distance H. Solving for the landing time gives a quadratic, not the simple flat-ground formula. Horizontal motion is still constant-velocity, and impact speed follows cleanly from energy conservation.
The single new idea is that the projectile does not finish at its launch height, so must remove the full launch-level rise AND an extra of cliff. Writing the vertical equation as and using the quadratic formula yields — the negative root is rejected because it corresponds to a time before launch. **Why energy is the shortcut for speed:** resolving the impact velocity into components and recombining works, but conservation of energy gives in one line, so — mass cancels and the angle never appears, because energy doesn't care about direction. **Limiting cases:** set and the time collapses to the familiar flat-ground and ; make huge and the launch angle barely matters — the stone essentially just falls off the cliff. **Connect it:** this is ordinary projectile motion with one honest bookkeeping change — the finish line sits below the start line, so you can no longer assume time-up equals time-down.
Assuming the flight is symmetric (time up = time down) or using the level-ground range formula R = v₀²sin(2θ)/g when the projectile lands BELOW the launch height. Those shortcuts only hold when the projectile returns to its launch level — off a cliff it keeps falling, so you must solve the quadratic y = H + v₀sinθ·t − ½gt² = 0 for the true flight time.