Rocket Motion Simulator

Scenario

A rocket fires its engines to launch straight up from rest with an upward acceleration of 5 m/s² for 10 seconds. After this time, the engines shut off, and the rocket freely falls straight back to Earth's surface.

5.0 m/s²
10.0 s
1.0x
Current Time: 0.00 s | Velocity: 0.00 m/s | Height: 0.00 m | Phase: Ready

Acceleration vs Time

Velocity vs Time

Answers to Problem Questions

Part C: Velocity after 10 seconds = 50.0 m/s

Part D: Height after 10 seconds = 250.0 m

Part E: At 10.1 seconds (after engines shut off), the rocket is in free fall. The acceleration is -9.8 m/s² (downward), which is the acceleration due to gravity.

Part F: At 10.1 seconds, the rocket is still moving upward because it has positive velocity. It won't reach maximum height until velocity becomes zero.

Time to Max Height: 15.10 s

Maximum Height: 377.55 m

Time to Land: 18.77 s

Understanding the Physics

Phase 1 (0-10s): Powered Flight

During powered flight, the rocket experiences constant upward acceleration of 5 m/s². The velocity increases linearly: v = at = 5t. After 10 seconds, v = 50 m/s. The height increases quadratically: h = ½at² = ½(5)(10)² = 250 m.

Phase 2 (10s-15.1s): Free Fall (ascending)

After engines shut off, only gravity acts on the rocket (a = -9.8 m/s²). The rocket continues upward but decelerates. Using v = v₀ + at, when v = 0: 0 = 50 + (-9.8)t, so t = 5.10 more seconds. Time to max height = 10 + 5.10 = 15.10 s.

Additional height gained: h = v₀t + ½at² = 50(5.10) + ½(-9.8)(5.10)² = 127.55 m. Total max height = 250 + 127.55 = 377.55 m.

Phase 3 (15.1s-18.77s): Free Fall (descending)

The rocket falls back to Earth from maximum height. Using y = y₀ + v₀t + ½at² with y = 0, y₀ = 377.55 m, v₀ = 0, and a = -9.8 m/s²: 0 = 377.55 + 0 + ½(-9.8)t². Solving: t = 8.77 s. Total time to landing = 10 + 8.77 = 18.77 s.