Resolving the Launch — Two Motions, One Clock
Send a cart along the floor at v₀cos θ and a bead up a mast at v₀sin θ, and the ball is always exactly where those two cross — for the whole flight, not just at the start. The amber ray shows where it would be with gravity switched off, and the ball hangs precisely ½gt² below it at every angle.
Bu simülasyon nasıl kullanılır
- Set the launch speed and angle and watch the two components split out
- Drag the time slider to move the cart, the bead and the ball together
- Follow the amber gravity-off ray and measure the gap down to the ball
- Load the worked example at 19.6 m/s and 30°, then try 60° for comparison
Nelere dikkat etmeli
- The ball sits at the crossing of the cart's position and the bead's height at every instant
- The cart moves at a steady rate for the whole flight while the bead slows, stops and returns
- The gap below the amber ray is ½gt² and depends only on the time, never on the angle
- A steeper launch gives a taller bead and a slower cart, for the same launch speed
- At 0° the bead never rises and the motion is a plain horizontal throw
Arkasındaki fizik
A launch velocity resolves into vₓ = v₀ cos θ and vᵧ = v₀ sin θ, and because the only force is vertical those two motions never interfere. Horizontally x = v₀ cos θ · t with no acceleration at all; vertically y = v₀ sin θ · t − ½gt². The first two terms together are simply the straight-line path the ball would follow with gravity off, so the whole effect of gravity is the single term ½gt² subtracted downwards — the same amount at every launch angle, since g does not know which way the ball was thrown.