A model rocket is launched from rest at ground level with a constant upward acceleration of 20 m/s ^2 . After 2 s , its engine cuts off and it moves freely under gravity. On its way down, when its downward velocity reaches 20 m/s , a parachute opens. The parachute provides a constant upward deceleration such that the rocket comes to rest exactly as it touches the ground. The magnitude of the deceleration provided by

Step-by-step solution

Stage 1 (Powered flight): Initial velocity u = 0 , upward acceleration a₁ = 20 m/s ^2 , time t = 2 s . Height reached h₁ = 1 2 a₁t^2 = 1 2 20 2^2 = 40 m . Velocity at engine cut-off v₁ = a₁t = 20 2 = 40 m/s (upwards). Stage 2 (Free fall under gravity): The rocket moves under gravity ( g = 10 m/s ^2 downwards) until its velocity becomes v₂ = -20 m/s (downwards). Using v₂^2 - v₁^2 = 2(-g)h₂ , where h₂ is the displacement from the engine cut-off point: (-20)^2 - (40)^2 = 2(-10)h₂ 400 - 1600 = -20h₂ -1200 = -20h₂ h₂ =