Answer :
The kinematics allows to find the result for the initial speed of the rocket is
v₀ = 2.23 m / s with a direction of θ = 54.1º
Kinematics studies the movement of bodies, finding relationships between position, velocity and acceleration.
The reference system is a coordinate system with respect to which all measurements must be made, in this case they indicate the accelerations and veloicties with respect to an x and y coordinate system,
The accelerations are aₓ = 5.10 m/s² and [tex]a_y[/tex] = 7.30 m/s², it also indicates the time that the motors are on t = 565 s, when the acceleration is finished the final speeds are vₓ = 3775 m/s and [tex]v_y[/tex] = 4816 m/s.
Let's use the relationship
vₓ = [tex]v_{ox}[/tex] + aₓ t
[tex]v_{ox} = \frac{v_x}{a \ t}[/tex]
Let's calculate
v₀ₓ = [tex]\frac{3775}{5.1 \ 565}[/tex]
v₀ₓ = 1.31 m / s
We perform the same calculation for the y axis
[tex]v_y = v_{oy} + a_y t \\v_{oy} = \frac{v_y}{a_y t}[/tex]
[tex]v_{oy} = \frac{4816}{ 7.3 \ 565 }[/tex]
[tex]v_{oy}[/tex] = 1.81 m / s
To find the velocity vector in the form of a module and direction we can use the Pythagoras' Theorem
v₀ = [tex]\sqrt{v_{ox}^2 + v_{oy}^2}[/tex]
v₀ = [tex]\sqrt{1.31^2 + 1.81^2}[/tex]
v₀ = 2.23 m / s
To find the direction we use trigonometry
tan θ = [tex]\frac{v_{oy}}{v_{ox}}[/tex]
θ = tan⁻¹ [tex]\frac{v_{oy}}{v_{ox}}[/tex]
θ = tan⁻¹ [tex]\frac{1.81}{1.31}[/tex]
θ = 54.1º
In conclusion with kinematics we can find the result for the initial speed of the rocket is
v₀ = 2.23 m / s with a direction of θ = 54.1º
Learn more here: brainly.com/question/13202575