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

Question

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 ? 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,?, g, and/or numerical factors Asked by

Explanation / Answer

find velocity at top of the incline

v^2 = v0^2 + 2 a L

v = sqrt(2 a L)

so vx = sqrt( 2 a L) cos theta

vy = sqrt(2 a L) sin theta

and x = L cos theta

y = L sin theta

at highest point vy = 0

vy^2 = v0y^2 + 2 a y

0 = 2 a L sin(theta)^2 - 2 * g*(y - L sin(theta))

y= L sin(theta) ( a sin(theta) + g)/g

now find time to vy = 0
0 = sqrt(2 a L) sin theta - g t
t = sqrt(2 a L) sin(theta)/g

x = Lcos theta + v0x t =Lcos theta + sqrt( 2 a L) cos theta * sqrt(2 a L) sin(theta)/g =Lcos theta + 2 a L sin(theta) cos(theta)/g

so r = <L cos theta + 2 a L sin(theta) cos(theta)/g, L sin(theta) ( a sin(theta) + g)/g>

b) so y = L sin theta
so
L sin theta = L sin theta + sqrt(2 a L) sin(theta)*t - 1/2*g*t^2
sqrt(2 a L) sin(theta) = 1/2 g t
t = 2 sqrt(2 a L) sin(theta)/g

now x direction
x = L cos(theta) + vx t = L cos(theta) + sqrt( 2 a L) cos(theta)*2 sqrt(2 a L) sin(theta)/g = L cos(theta) + 4 a L sin theta costheta/g

so r = <L cos(theta) + 4 a L sin theta costheta/g, L sin theta>

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