First use v^2 = u^2 + 2as with a = 10 , u =0, s = 500 and find v which is the velocity at 500 m .
Now find the time taken to reach 500 m height say t1 using
v = 0 + 10 . t1
The rocket still moves up (though the engines are switched off) due to inertia but slows down till its vel becomes zero. Let the time taken from 500 m upwards till the vel becomes zero be t2. Use
0 = v - g . t2 to find t2.
Now use 0^2 = v^2 - 2 . g .h to find the height (say h) it rises above 500m. The total height reached by the rocket before it starts falling is thus (500 + h). Now onwards it is free fall downwards from this height. Let the time taken for this be t3. Find t3 using
(500 + h) = 0 . t3 + 0.5 g (t3)^2.
The time it is in the air is thus t1 + t2 + t3.
This was worked out without the given max height of 1010 only to show that it can be done even without that bit of data. According to the equations the max height works out to be 1010.2. Using the given data and g = 9.8 m/s^2, the total time works out to be 34.56 secs.
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Last edited by physicsquest; Apr 8th 2010 at 11:38 PM.
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