Kinematics: Rocket/Free Fall Problem

  • Thread starter Thread starter beauch1322
  • Start date Start date
  • Tags Tags
    Fall Kinematics
AI Thread Summary
The discussion revolves around solving a kinematics problem involving a rocket's flight with constant acceleration followed by free fall. The rocket accelerates for 5 seconds before entering free fall, reaching a maximum height of 1200 meters. Key equations include those for final velocity and displacement, which help determine the rocket's acceleration and total time of flight. The user struggles with substituting variables effectively but receives guidance on using the final velocity from the first phase as the initial velocity for the second phase. The conversation emphasizes applying kinematic equations systematically to solve the problem.
beauch1322
Messages
2
Reaction score
0

Homework Statement


A one stage booster rocket has a Xo of 0m and a Vo of 0 m/s. The rocket takes off with a constant acceleration until 5 seconds when the boosters turn off and goes into free fall. The rocket is set to explode the instant it hits it's max height. The max height is 1200m.

What is the acceleration for the first 5 seconds?
What is the Vf for the first 5 seconds?
What is the total time of flight?


Homework Equations


Xf=Xo+Vo(t)+(.5)a(t^2)
Vf=Vo+at
(Vf^2)=(Vo^2)+2a(Xf-Xo)



The Attempt at a Solution



1st Part
Xo1=0m
Xf1=Xo2=
Vo1=0m/s
Vf1=Vo2=
a1=
t1=5s

Part 2
Xo2=Xf1=
Xf2=
Vo2=Vf1=
Vf2=0
a2=-9.8 m/s^2
t2=

Also we know: Xf1+Xf2=1200m

I kept trying to get to an equation where I could boil it down to one variable but I can't seem to do it... I have used a lot of substitution, please help...


THANK YOU FOR YOUR TIME!
 
Physics news on Phys.org
You've got a good start. For part 1 you can simply say vf=a*t. Now the problem is really what I'm sure you've done over and over, just a kinematic problem with an initial velocity and height. Use that "final velocity" as an initial velocity in problem two. Then in problem two you can use the distance kinematic equation, and you can get time from the velocity kinematic equation. So on and so forth. Try it out with some confidence and come back if you're stuck.
 
got it. I used a m/s vs t graph and solved for Vf1... 1200= (5*Vf1)/2 + (Vf1*(Vf1/9.8))/2
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top