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September 20, 2014

September 20, 2014

Posted by **jon** on Monday, November 23, 2009 at 9:15am.

How can I derive the equation for the trajectory of the probe during the time the rocket is fired using the formula y = a function of x and compute the final speed of the rocket at the end of the 1.5 s?

- physics -
**MathMate**, Monday, November 23, 2009 at 12:35pmThrust is in force units. For all intents and purposes, there is a force in the y-direction of 1800 N for 1.5 seconds.

The motions in the x- and y-directions being linearly independent, we can process them separately and relate the motion using the parameter t=time.

vx(t)=12 m/s (constant)

force in y-direction, F = 1800 N

mass, m = 2000 kg

Since Force F=ma, where a = acceleration

acceleration=F/m=1800/2000 m/s²

=0.9 m/s²

Initial velocity in the y-direction = 0

therefore

vy(t)=0+at=0.9t (0≤t≤1.5)

The trajectory is given by the parametric formulae vx(t) and vy(t) in terms of time, t.

The speed at any time t within the firing period (1.5 s) can be obtained by the vectorial sum of the two velocities, vx(t) and vy(t):

Speed for 0≤t≤1.5

S(t)=sqrt(vx(t)²+vy(t)²)

=sqrt(12²+(0.9t)²)

Speed at the end of the rocket firing period

=S(1.5)

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