Initial value problem: dropping a ball
About this app
h″ = −g with a falling ball: integrating twice leaves a family of parabolas with C₁ and C₂. With no condition many balls fall, with only h(0) still a family; only h(0) and h′(0) pick exactly one path. Draggable initial height and velocity (start tangent), boundary value problem with a target after t₁, Earth/Mars/Moon, working with numbers.
With explain mode: an AI tutor walks you through the app, points out what matters in the scene and answers your questions.
Why a second-order ODE needs two initial values
Without air drag the height of a falling body obeys h″(t) = −g. Integrating twice gives h(t) = −½gt² + C₁t + C₂: a family of parabolas with two free constants. An initial value problem prescribes as many conditions at one point as the order. The initial height h(0) = h₀ fixes C₂, the initial velocity h′(0) = v₀ fixes C₁. Only both together determine exactly one solution.
What the app shows
On the left there is a drop tower with a height scale, on the right the curve h(t). You choose what is given: nothing, only the height, only the velocity, both, or a target at a later time. For every remaining freedom ghost balls fall along, and their parabolas appear in the curve card. The start tangent shows h′(0) as a slope. The Working panel walks through integrating and substituting and gives impact time, impact speed and the highest point.
What you can try
Drag the ball to a new height and the arrow tip for the initial velocity. Switch from “h(0) and h′(0)” to “only h(0)” and watch many balls fall again. In the boundary value problem you require the ball to be at height h₁ after t₁ seconds; the app computes the matching initial velocity. On Mars and the Moon the same equation runs with a smaller g, and the strobe trail shows the steadily growing speed.
Frequently asked questions
- What is an initial value problem?
- An initial value problem consists of an n-th order differential equation and n conditions at the same point, such as the function value and its derivatives at t = 0. These conditions fix the free constants of the general solution and pick one solution.
- How do you solve h″ = −g for free fall?
- Integrating once gives h′(t) = −gt + C₁, a second time h(t) = −½gt² + C₁t + C₂. With h(0) = h₀ we get C₂ = h₀, with h′(0) = v₀ we get C₁ = v₀. So h(t) = h₀ + v₀t − ½gt².
- What is the difference between an initial and a boundary value problem?
- In an initial value problem all conditions sit at one point, for example height and velocity at release. In a boundary value problem they sit at two or more points, for example starting at 30 m and being on the ground after 3 s. Boundary value problems can be unique, unsolvable or ambiguous.
- What can I adjust in the app?
- You choose the given conditions, drag the initial height and velocity, set target time and height for the boundary value problem, and switch between Earth, Mars and the Moon. An explain mode demonstrates the key steps.
Subject: Differential equations
More from Differential equations
A VisuApp by heyprof: interactive, free in your browser, no sign-up. What is a VisuApp? · All apps