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redworld33

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A test rocket is launched, starting on the ground, from rest, by

accelerating it along an incline with constant acceleration a. The incline has length

L, and it rises at θ degrees above the horizontal. At the instant the rocket leaves the

incline, its engines turn off and it is subject only to gravity, g≡+9.81m/s2. (Air resistance

can be ignored). Taking the usual x-y coordinate system, with an origin at the top edge

of the incline...

(a) What is the position vector when the rocket is at its highest point?

(b) What is the position vector when the rocket is on its way back down and once again at

the same height as the top edge of the incline?

Your symbolic answer should only

depend on a, L, θ, g, and/or numerical factors.

I am so lost on this problemMy attempt:

Height of the incline = LSinθ

Angle: θ=arcsin(Height/L)

Horizontal velocity after rocket has finished accelerating up the incline = cosθ * sqrt(2aL)

Vertical velocity after the rocket has finished accelerating up the incline = sinθ * sqrt(2aL)

accelerating it along an incline with constant acceleration a. The incline has length

L, and it rises at θ degrees above the horizontal. At the instant the rocket leaves the

incline, its engines turn off and it is subject only to gravity, g≡+9.81m/s2. (Air resistance

can be ignored). Taking the usual x-y coordinate system, with an origin at the top edge

of the incline...

(a) What is the position vector when the rocket is at its highest point?

(b) What is the position vector when the rocket is on its way back down and once again at

the same height as the top edge of the incline?

Your symbolic answer should only

depend on a, L, θ, g, and/or numerical factors.

I am so lost on this problemMy attempt:

Height of the incline = LSinθ

Angle: θ=arcsin(Height/L)

Horizontal velocity after rocket has finished accelerating up the incline = cosθ * sqrt(2aL)

Vertical velocity after the rocket has finished accelerating up the incline = sinθ * sqrt(2aL)

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