Calculating Rocket's Initial Velocity with 45-Degree Angle

AI Thread Summary
To calculate a rocket's initial velocity fired at a 45-degree angle, the horizontal displacement of 22 meters over 1.9 seconds indicates a constant horizontal velocity. Using the formula for horizontal motion, the horizontal component of the velocity can be determined. The participant initially calculated both the horizontal and vertical initial velocities as 1.2 m/s, but clarification on the projectile motion principles is needed. The horizontal component remains constant, while the vertical component is influenced by gravity. Understanding these concepts is crucial for accurate calculations in projectile motion.
mitchy_boy
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Homework Statement


a rocket is fired at an angle of 45degrees from the horizontal, if it travels 22m(horizontally) for a total time of 1.9 secs, what is its intitial veritical and horizontal velocity?


Homework Equations


i tried using s=ut+.5at^2


The Attempt at a Solution


i came up with 1.2 for both horizontal and initial vertical velocity's

please help I am confused :S
 
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Hi mitchy boy, welcome to PF.
In the projectile motion horizontal component of the velocity remains constant. In the problem horizontal displacement and time is given. From that find the horizontal component of the velocity.
 

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