A toy rocket is launched upward with a net acceleration of 10m/sq.sec. for 3 sec. It then slows at the rate of 10m/sq.sec. until it reaches its maximum altitude. How high does it go?
by 
bevis21
ANSWERS: 1
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The first part is symmetrical to the second. Equation for distance travelled under acceleration is s = u t + (1/2) a t^2 where u is the initial speed (0 probably). So s = 45m for the first half and the second part is symmetrical.
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