Hilton T. answered • 10/06/15
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Former university lecturer with over 12 years experience
This question cannot be answered without additional information. We need to know the initial velocity of the rocket, and from what height it is launched. If we assume this is zero, and it is launched from the ground, then the rocket reaches a height of 1/2 at^2 = 1/2 x 4 x 6^2 = 72 m while it is accelerating. After 6 seconds its velocity is 6 x 4 = 24 m/s. After the 6 seconds, the rocket's acceleration is -g = -10 m/s^2. The additional time taken by the rocket to return to the ground is determined by the following.
Vi = 24 m/s a = - 10 m/s ^2 d = -72 m t = ?
Using, d = vit + 1/2 at ^2 -72 = 24t+ 1/2 x (-10) x t^2 5t^2 -24 t -72 =0 Solve for t using the quadratic formula gives t = 6.9 s.
Total time in air = 6 + 6.9 = 12.9 s
Of course this answer is based on the assumptions mentioned above.
