domingo, 5 de enero de 2020

mecànica el trampolín parabólic

qgH + (-1)·qgy + (-1)·(m/2)·d_{t}[y]^{2} = qgy + (m/2)·d_{t}[y]^{2}


y(t) = (-1)·(1/2)·( (qg)/m )·t^{2} + (H/2)


d_{t}[y(t)] = (-1)·( (qg)/m )·t


d_{t}[y(t)] = 0 <==> y(t) = (H/2)


d_{t}[y(t)] = a <==> y(t) = (-1)·(1/2)·( m/(qg) )·a^{2} + (H/2)


0 = (-1)·(1/2)·( (qg)/m )·t^{2} + (H/2)


(H/2) = (1/2)·( (qg)/m )·t^{2}
t = ( ( (Hm)/(qg) ) )^{(1/2)}


qg(H/2) = (m/2)·d_{t}[x]^{2}


x(t) = ( ( (qgH)/m ) )^{(1/2)}·t


abast:
x(( ( (Hm)/(qg) ) )^{(1/2)}) = ( ( ( (qgH)/m ) )^{(1/2)} )·( ( ( (Hm)/(qg) ) )^{(1/2)} )
x(( ( (Hm)/(qg) ) )^{(1/2)}) = H


trayectoria de vuelo:
y = (-1)·( 1/(2H) )·x^{2}+(H/2)


y = (H/2) <==> x = 0


0 = (-1)·( 1/(2H) )·H^{2}+(H/2) <==> ( x = H or x = (-H) )


trayectoria de la rampa:
y = ( 1/(2H) )·x^{2}+(H/2)


y = (H/2) <==> x = 0


H = ( 1/(2H) )·H^{2}+(H/2) <==> ( x = H or x = (-H) )

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