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A 1500 kg weather rocket accelerates upward at 10m/s2. It explodes 2.0 s after liftoff and breaks into two fragments, one twice as massive as the other. Photos reveal that the lighter fragment traveled straight up and reached a maximum height of 530 m. Part A What was the speed of the heavier fragment just after the explosion? Express your answer with the appropriate units.

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lcmendozaf

Answer:

[tex]V_H=-20.45m/s[/tex]

Explanation:

By kinematics:

[tex]V_R=V_o+a*t=0+(10m/s^2)*(2s)=20m/s[/tex]

The rocket's height at this point is:

[tex]V_R^2=V_o^2+2*a*H_R[/tex]

[tex]H_R=20m[/tex]

On the lighter fragment:

[tex]V_f^2=V_L^2-2*g*H_L[/tex]  where Vf=0. Solving for VL:

[tex]V_L=\sqrt{2*g*H_L}=100.99m/s[/tex]

By conservation of the momentum:

[tex]3*m*V_R=2*m*V_H+m*V_L[/tex]

Solving for [tex]V_H[/tex]:

[tex]V_H=\frac{3*V_R-V_L}{2}=-20.45m/s[/tex]

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