Calculate T for Maximum Altitude: 1D Kinematics Problem Solution

[David112234]

Homework Statement

[/B]
During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be launched from rest from the earth’'s surface and is to reach a maximum height of 990 m above the earth'’s surface. The rocket’'s engines give the rocket an upward acceleration of 16.0 m/s2 during the time T that they fire. After the engines shut off, the rocket is in free fall. Ignore air resistance.

What it asks for : What must be the value of T in order for the rocket to reach the required altitude?

Homework Equations

basic kinematics formulas
X= Xo + VoT+1/2AT^2
V=Vo+AT
V^2=Vo^2+2A(X-Xo)

o denotes initial

The Attempt at a Solution

| X3= 990 V3=0
|
|
| X2= V2= A=-9.8
|
|
|
| X1=0 V1=0 A=16.0 m/s

X3=X2+V2T-1/2 9.8T^2
X2=X1+V1T+.5*16*T^2
X2=0+0T+.5^16*T^2
X2=8T^2

plugging pack into original equation:
X3=8T^2 +V2T -4,9 T^2
combining like terms
X3= 3.1T^2 + V2T

finding V2:
V2=V1+16T
V2=0+16T
V2=16T
plugging back in

X3= 3.1T^2 + 16T
X3=990
990=3.1T^2 + 16T

it is a quadratic equation I graphed it to find the X value, I got 20.65 , but it is wrong, where Did I go wrong?

 

[Fightfish]

The 'T's that appear here

X3=X2+V2T-1/2 9.8T^2

and

X2=X1+V1T+.5*16*T^2

are not the same.
The first is the time it takes for the rocket to reach zero velocity after the engines were shut off, while the second is the boost time during which the engines were turned on.

 

[David112234]

The 'T's that appear here

and
are not the same.
The first is the time it takes for the rocket to reach zero velocity after the engines were shut off, while the second is the boost time during which the engines were turned on.

So what alternative method should I use?

 

[Fightfish]

There is no need for an alternative method; your general approach is okay - just that you need to be careful when you keep track of the two different times. To proceed, you have to find out how the time $t_{2}$ in $x_{3}=x_{2}+ v_{2}t_{2} -\frac{1}{2} g {t_{2}}^2$ is related to the boost time $T$.
(Hint: Use $v=v_{0}+at$)

 

[David112234]

There is no need for an alternative method; your general approach is okay - just that you need to be careful when you keep track of the two different times. To proceed, you have to find out how the time $t_{2}$ in $x_{3}=x_{2}+ v_{2}t_{2} -\frac{1}{2} g {t_{2}}^2$ is related to the boost time $T$.
(Hint: Use $v=v_{0}+at$)

V3=V2+AT
V2=16t
V3=16t-9.8T
V3=0
0=16t-9.8T
-16t=-9.8T
16/9.8t = T

8t²+16t²-4.9T²
8t²+16t²-4.9(16/9.8 t)²

990= 10.93 t²
90.5=t²
√90.5 =t
t= 9.513

the correct answer is 6.86.

 

[Fightfish]

the correct answer is 6.86.

Yup, that's what I got.

This line is not correct:

$$x_{3} = 8t^2+16t^2-4.9T$$

It should be
$$x_{3} = x_{2} + v_{2}t - \frac{1}{2}g t^2 = \frac{1}{2}a T^{2} + (aT)t - \frac{1}{2}g t^2$$
Substitute $t = (a/g)T$ as you've derived above, and you should be done.

(Note: I realized the meaning of t and T I've used here is opposite of what you have in that post, but you should be able to make the connection :p)