Calculating Rocket's Initial Velocity with 45-Degree Angle

[mitchy_boy]

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

 

[rl.bhat]

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.

 

[kerimek]

I would approach this problem by first defining the variables and using the appropriate equations to solve for the initial velocity. In this case, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Since the rocket is traveling at an angle of 45 degrees, we can break down its motion into horizontal and vertical components. The horizontal component of the initial velocity (ux) is equal to 22m (the distance traveled horizontally) divided by 1.9s (the total time), giving us a value of 11.6 m/s.

For the vertical component (uy), we can use the equation uy = u*sin(θ), where θ is the angle of 45 degrees. This gives us a value of 11.6 m/s * sin(45) = 8.2 m/s.

Therefore, the initial velocity of the rocket can be calculated using the Pythagorean theorem: u = √(ux^2 + uy^2) = √(11.6^2 + 8.2^2) = 14.2 m/s.

So, the initial horizontal and vertical velocities of the rocket are 11.6 m/s and 8.2 m/s, respectively. It is important to note that these values are only accurate if the acceleration is constant and there are no external forces acting on the rocket.