Answer to Question #140728 in Physics for Derek

Question

Motion of the rocket is given by the following equation:

x=5+5t+7t²+10t³ ; 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. 

Round your answer to the nearest tenth of a meter.

Answer 1

After the piece detaches from the rocket, it starts moving under the influence of gravity. So, find the initial height of the object:

$$x=h_0=5+5(5)+7(5)^2+10(5)^3=1455\text{ m}.$$

Find the equation of velocity of the rocket (and piece before it detached), which is derivative of height by time:

$$v=x'=5+14t+30t^2,\\ v=825\text{ m/s}.$$

If the rocket is moving straight vertically, the piece's height will decrease:

$$h=x-v_0t-\frac{gt^2}{2}=\\\space\\ =1455+825\cdot2-\frac{9.8\cdot2^2}{2}=3085.4\text{ m}$$

Meanwhile after 7 seconds after launch, the height of the rocket is 3813 m.