Solving for Launch Speed and Height of a Toy Rocket

AI Thread Summary
The discussion revolves around calculating the launch speed and maximum height of a toy rocket that passes a window 2.2 m high, with its sill 9.0 m above the ground. The rocket takes 0.14 seconds to travel the height of the window, leading to questions about the correct method to find its initial velocity. Participants suggest using kinematic equations, specifically Vf² - Vi² = 2a x, to relate the final and initial velocities and the distance traveled. There is confusion among users regarding the application of these equations, prompting requests for clearer explanations and guidance. The conversation emphasizes the need for algebraic manipulation to solve for the unknown variables effectively.
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A toy rocket moving vertically upward passes by a 2.2 m high window whose sill is 9.0 m above the ground. The rocket takes 0.14 m/s} to travel the 2.2 m height of the window.

What was the launch speed of the rocket? Assume the propellant is burned very quickly at blastoff.

How high will the rocket go?
 
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bump please help
 
homeworkboy said:
A toy rocket moving vertically upward passes by a 2.2 m high window whose sill is 9.0 m above the ground. The rocket takes 0.14 m/s} to travel the 2.2 m height of the window.

What was the launch speed of the rocket? Assume the propellant is burned very quickly at blastoff.

How high will the rocket go?

What do you think you should do to solve the problem?
 
well since it says it tokk .14 sec to travel 2.2 m then velocity (v) = 2.2/.14 then we get the final velocity...we know a = -9.8 m/s2...and X = 2.2+9 =11.2m...so we find initial velocity but it says that its the wrong answer

so if u teach me the method it would be great...
 
homeworkboy said:
well since it says it tokk .14 sec to travel 2.2 m then velocity (v) = 2.2/.14 then we get the final velocity...we know a = -9.8 m/s2...and X = 2.2+9 =11.2m...so we find initial velocity but it says that its the wrong answer

so if u teach me the method it would be great...

You need to use the relationship that Vf2 - Vi2 = 2 a x where a is 9.8 m/s2

You are given 2 distances here. Choose one formula from launch.

Vo2 - Vb2 = 2 a x ... where Vb is V bottom and x is height to bottom of window which is 9

You also know that since gravity is slowing things down that
Vb = Vt + .14 (9.8) where Vt is velocity at the top of the window.

Then you can make use of the other equation to the top of the window:
Vo2 - Vt2 = 2 a x ... where Vt is V top and x is height to top of the window which is 11.2

3 equations. 3 unknowns. Solve.
 
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I dint understand your explanation...is there anyone else who can help me..
 
bump...anyone
 
homeworkboy said:
I dint understand your explanation...is there anyone else who can help me..

What have you tried? You have some of the values. And you have 3 equations that relate the variables you don't have to each other.

You need to use algebra to solve the equations and eliminate your unknowns.
 
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