# 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

- NCSLv 71 month agoFavorite 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

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- Anonymous1 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)