Finding the Maximum Time for a Rocket's Flight Using Kinematics

[sherlockjones]

A rocket initially at rest accelerates with constant net acceleration B from t = 0 to t = T1 at which time the fuel is exhausted. Neglect air resistance. If the rocket's net acceleration, B, is equal to 1.0g, find an expression for the total time T_{max} (from liftoff until it hits the ground).


So T_{max} = T_{1} + t

\frac{1}{2}BT_{1}^{2} + BT_{1}t - \frac{1}{2}gt^{2} = 0

I know that t = \frac{BT_{1}}{g}


What do I do from here? I got T_{max} = 2T_{1} = 2 t


Thanks

 

[radou]

Looks okay, since, when the rocket is left without any fuel, its motion is a free fall with y(t) = yo + vo t - 1/2 g t^2, where yo is the well-known height yo = y(T1) = 1/2 B T1^2 = 1/2 g T1^2, and v0 = BT1. You're on the right track. Now just solve for t, and plug it into Tmax = T1 + t.

 

[sherlockjones]

So is the equation \frac{1}{2}gt^{2} + gt^{2} - \frac{1}{2}gt^{2}?

 

[radou]

The equation is \frac{1}{2}gT_{1}^{2} + gT_{1}t - \frac{1}{2}gt^{2} = 0, as you already wrote. Now solve for t.

 

[sherlockjones]

If B = g how do we get t^{2} - 2T_{1}t - T_{1}^{2} = 0?

I factored the equation: g(\frac{1}{2}T_{1}^{2} + T_{1}t - \frac{1}{2}t^{2}) = 0. I guess they used the relation that t = T_{1} and multiplied both sides by 2?

 

[radou]

If B = g how do we get t^{2} - 2T_{1}t - T_{1}^{2} = 0?

I factored the equation: g(\frac{1}{2}T_{1}^{2} + T_{1}t - \frac{1}{2}t^{2}) = 0. I guess they used the relation that t = T_{1} and multiplied both sides by 2?

Again, solve the equation (i.e. find the roots of the parabola) \frac{1}{2}gT_{1}^{2} + gT_{1}t - \frac{1}{2}gt^{2} = 0 for t. It is the only unknown.