When it is 141 m above the ground, a rocket traveling vertically upward at a con

Question

When it is 141 m above the ground, a rocket traveling vertically upward at a constant 8.90 m/s relative to the ground launches a secondary rocket at a speed of 12.0 m/s at an angle of 59.0 above the horizontal, both quantities being measured by an observer sitting in the rocket. Air resistance should be ignored.

Part A:

Just as the secondary rocket is launched, what are the horizontal and vertical components of its velocity relative to the astronaut sitting in the rocket?

Enter your answers numerically separated by a comma.

Part B:

Just as the secondary rocket is launched, what are the horizontal and vertical components of its velocity relative to Mission Control on the ground?

Hint: You need to use Galileo's equation that we discussed last week to find the initial velocity with respect to the ground. Frame A is the ground, Frame B is the first rocket and the object of interest P is the secondary rocket.

Enter your answers numerically separated by a comma.

Part C: Find the initial speed of the secondary rocket as measured by Mission Control. Speed is the magnitude of the velocity vector of the secondary rocket with respect to the ground.

Part D: Find the launch angle of the secondary rocket as measured by Mission Control.

Part E: What maximum height above the ground does the secondary rocket reach? This is a standard projectile motion problem but all quantities used in the formulas should be with respect to the reference frame, which in this case is the ground. So the initial position of the secondary rocket with respect to the ground is the launch height. Max height occurs when the y component of the velocity with respect to the ground is zero. Acceleration on the secondary rocket is -g. Use the most convenient of the kinematic equations - the one that does not involve time.

Explanation / Answer

Vr = 8.9 j^ m/s

Part A :-
Velocity relative to the astronaut sitting in the rocket -
Vsr = 12.0 * cos(59) i^ + 12.0 * sin(59) j^

Horizontal component, Vx = 6.18 m/s
Vertical Component, Vy = 10.28 m/s

Part B:-
Velocity relative to Mission Control on the ground :-
Vsr wrt ground = 12.0 * cos(59) i^ + (12.0 * sin(59) +8.9) j^ m/s

Horizontal component, Vx = 6.18 m/s
Vertical Component, Vy = 19.18 m/s

Part C:-
Speed, |V| = sqrt(Vx^2 + Vy^2) m/s
Speed, |V| = sqrt(6.18^2 + 19.18^2) m/s
Speed, |V| =  20.15 m/s

Part D:-
Launch Angle = tan^-1(19.18/6.18)
Launch Angle = 72.14o

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