Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 month ago

Rocket Problem Help?

The motion of the rocket is given by the following equation:

x=5+5t+7t^2+10t^3 ; where t is in seconds, x is in meters.

At 5s a small object detaches itself from the body of the rocket.

Find the height of the object two seconds after it was detached from the rocket.

2 Answers

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  • NCS
    Lv 7
    1 month ago
    Favorite Answer

    Given: x(t) = 5 + 5t + 7t² + 10t³

    Then v(t) = dx/dt = 5 + 14t + 30t²

    at t = 5, x(5) = 1455 m

    and v(5) = 825 m/s

    by simple substitution

    Two seconds after detachment,

    h = x + v*t - ½gt² = 1455m + 825m/s*2s - ½*10m/s²*(2s)²

    using a nice round number for g.

    h = 3085 m

    If you find this helpful, please select Favorite Answer!

  • Anonymous
    1 month ago

    after 5 sec

    h = 5+5t+7t^2+10t^3 = 5+5*5+7*5^2+10*5^3 = 1455 m 

    V is the first derivative of the space

    V = 5+7*2*5+10*3*5^2 = 5+70+ 750 = 825 m/sec 

    after 2 sec from the object detachement

    H = h+V*2-g/2*2^2 = 1455+825*2- 4.903*2^2 = 3090 m (3 sign. digits)

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