Find the landing event from the actual initial height

Begin with a measurement question

The familiar same-height projectile range formula has a condition: launch and landing heights must match. The supplied virtual point starts 2 m above the ground. Resolve its initial velocity into horizontal vx and vertical vy, with upward positive, then share one elapsed time between both component equations. Horizontal velocity remains constant only because air resistance and other horizontal forces are excluded.

Build the algebraic model

The trajectory is x=vxt and y=2+vyt−5t² with g=10 m/s². Landing is the first future solution of y=0. The positive root is [vy+√(vy²+40)]/10; the other root is negative for these controls and does not represent a future event. Multiply the accepted time by vx for the horizontal landing coordinate.

Check a concrete implication

At a future apex, vertical velocity becomes zero while vertical acceleration stays −10 m/s². Horizontal velocity remains nonzero. If vy is initially downward, the maximum height during the modeled future flight is the starting 2 m; an extrapolated earlier vertex is not part of that flight. End the plotted path at contact rather than extending the ideal parabola through the ground.

Challenge the evidence boundary

For vx=8 m/s and vy=4 m/s, flight time is approximately 1.14833 s and horizontal landing coordinate 9.18665 m. Vertical impact velocity is about −7.48331 m/s. The two components determine impact speed, but the model does not describe contact forces, rebound, finite-size objects or practical launch safety. All activity remains virtual.

Worked original case

At vy=0, landing time is√0.4≈0.632456 s for any horizontal velocity. Changing vx changes horizontal distance, not fall time under these independent-component assumptions.

Predict and investigate

Does greater horizontal speed keep this ideal point above the ground longer? Compare vx=4 and12 at the same vy, then vy=−2, 0, 4 at vx=8. Retain the accepted root, endpoint and future maximum-height classification. State an observation that would challenge the no-horizontal-force assumption.

Changed numerical practice

Try the changed case before feedback

A virtual point begins 5 m above ground with vy=0 and vx=6 m/s under g=10. Find horizontal landing distance in metres.

Numerical result in the requested unitsCompare this result

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Reflect and source-check

Why is zero vertical velocity at the apex not evidence for zero gravity?

Original explanations, synthetic records and models; checked 30 October 2026 against OpenStax College Physics 2e. The guided calculation is algebra-based; no calculus prerequisite is assumed. All activity is virtual. No physical launch, drop, water crossing, machinery operation or medical decision is requested. A public answer or saved capture does not verify independent mastery or actual measurement technique.