Speed of a rocket using energy methods

[BrainMan]

Homework Statement

A rocket is launched such that when the fuel is exhausted, the rocket is moving with a speed of V_o at an angle of 37° with the horizontal and at an altitude h. (a) Use energy methods to find the speed of the rocket when its altitude is h/2. (b) Find the x and y components of velocity when the rockets altitude is h/2. Use the fact that V_x=V_xo = constant (since ax-0) and the result from a.

Homework Equations

The Kinetic and Potential energy equations

The Attempt at a Solution

I did
1/2mv²+mgy= E
and then I solved for v and got
v= (2Egy)^1/2
the correct answer is (a) v = (v₀² + gh)^1/2
(b) v_x= .799v_o, v_y= - (- 0.362v_o² +gh)^1/2
I wasn't really sure how to approach this type of problem

 

[vela]

Homework Statement

A rocket is launched such that when the fuel is exhausted, the rocket is moving with a speed of V_o at an angle of 37° with the horizontal and at an altitude h. (a) Use energy methods to find the speed of the rocket when its altitude is h/2. (b) Find the x and y components of velocity when the rockets altitude is h/2. Use the fact that V_x=V_xo = constant (since ax-0) and the result from a.

Homework Equations

The Kinetic and Potential energy equations

The Attempt at a Solution

I did
1/2mv²+mgy= E
and then I solved for v and got
v= (2Egy)^1/2

How did you manage to get that? It looks like you didn't do the algebra correctly.

the correct answer is (a) v = (v₀² + gh)^1/2
(b) v_x= .799v_o, v_y= - (- 0.362v_o² +gh)^1/2
I wasn't really sure how to approach this type of problem

 

[BiGyElLoWhAt]

in a you solved for v wrong. You're attempting to do what they did, you just made a mistake that I think you can find.
When they say the speed at h/2, do they mean on the way up or down?

 

[BrainMan]

I'm not sure and what could I do with that information?

 

[haruspex]

1/2mv²+mgy= E

You have two instances where you can apply that equation: at the point where the fuel is exhausted, and again when it has descended to height h/2. Which term is the same at both points? What equation does that give you?

 

[BrainMan]

Is it that they both have the same amount of energy? So would you get something like 1/2mv²+mgy=1/2mv²+mgy?

 

[haruspex]

Is it that they both have the same amount of energy? So would you get something like 1/2mv²+mgy=1/2mv²+mgy?

Yes, plugging in the appropriate values for v and y each side.

 

[BrainMan]

OK I get it but I am having trouble finding the y component. I did sin 37 to find the y component but it looks like in the answer they gave they have that number squared - (0.362vo2 +gh)1/2. Why is sin theta squared in this equation?

 

[SammyS]

OK I get it but I am having trouble finding the y component. I did sin 37 to find the y component but it looks like in the answer they gave they have that number squared - (0.362vo2 +gh)1/2. Why is sin theta squared in this equation?

What is $\ (v_0)_y\ ?\$ Isn't sin(θ₀) a factor of this?

 

[BrainMan]

The initial speed in the y direction is Vo* sin(37). So shouldn't the equation be .601Vo instead of .362vo?

 

[haruspex]

The initial speed in the y direction is Vo* sin(37). So shouldn't the equation be .601Vo instead of .362vo?

If a speed is 0.601v, what is the square of that speed?

 

[BrainMan]

OK I see how they got that answer. Thanks everyone!