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.