A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.4 m/s at ground level. The engines then fire, and the rocket accelerates upward at 3.90 m/s2 until it reaches an altitude of 1130 m. At that point its engines fail, and the rocket goes into free fall, with an acceleration of −9.80 m/s2. (You will need to consider the motion while the engine is operating and the free-fall motion separately.)
What is its velocity just before it hits the ground?

Question

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.4 m/s at ground level. The engines then fire, and the rocket accelerates upward at 3.90 m/s2 until it reaches an altitude of 1130 m. At that point its engines fail, and the rocket goes into free fall, with an acceleration of −9.80 m/s2. (You will need to consider the motion while the engine is operating and the free-fall motion separately.)
What is its velocity just before it hits the ground?

anonymous · Answer

Dumbest catapult in the world...

In any case, v_0=80.4, a=3.9, x-x_0=1130; so, you can solve for t in x-x_0=v_0t+(1/2)at^2 using the quadratic formula.

Now, take that t and solve for v in v=v_0+at.

Now, for the free fall part. The a=-9.8. x-x_0=0-1130, v_0 is given by v in the previous part, so you can use x-x_0=v_0t+(1/2)at^2 again to solve for t. Then plug that t into v=v_0+at to solve for v, and voila, you have your answer.

anonymous · Answer

Is there a part you're having difficulty with? If I wasn't clear, please tell me.

anonymous · Answer

yes, i can't understand your symbols V_0 and x-x_0 ? i translated those into V_0= Vi ( initial) and x-x_0 = Height.. 
can u write the formula with those symbols because something goes wrong with me .. i get 7.2= t^2+t  :S

anonymous · Answer

Well, I won't give you a numerical answer--that's up to you. I'll just help you along. I guess my writing is undecipherable, so give me some time to type something up that is easier to read.

anonymous · Answer

sure, and i don't want the numerical answer because i won't learn anything that way :)

anonymous · Answer

Yuck, I was typing something up and it got deleted by the server reset. I'm feeling so depressed right now.

anonymous · Answer

le sigh... if you don't mind waiting for someone who's better at LaTeX, it might make things clearer

anonymous · Answer

First, let's track the motion upwards, $$x  = x_i + v_i t + {1 \over 2} a t^2$$It says that the rockets fire at ground level, therefore $x_i =0$. The catapult gives an initial velocity, therefore $v_i = 80.4$. Additionally, $a = 3.2$ and $x =1130$. We need to solve this equation for time. 

With the calculate time, we can determine the velocity at the point when the rockets burn out. $$v_b = a t$$

Now we need to find the distance the rocket continues to travel upwards after the rockets burn out and before it begins to fall back to earth. This can be found from $$0 = v_b^2 - 2 g \Delta x$$We need to solve for $\Delta x$
The height of the rocket when it begins to free fall is$$x_{ff} = 1130 + \Delta x$$Then we can find the velocity of the rocket when it hits the ground from the following expression. $$v_g^2 = 2 g x_{ff}$$

anonymous · Answer

i cant find $$\Delta X$$.. thats my problem..

anonymous · Answer

Why not?

anonymous · Answer

i guess my brain exploded or something, i've been solving physics and calculus all day long.. and now this is the last of them and i need to get some sleep. i'm not functioning correctly :)

anonymous · Answer

$$\Delta x = {v_b^2 \over 2g}$$

anonymous · Answer

ohh, i totally missed that equation you wrote earlier.. silly me :)