A rocket is fired vertically with an upward acceleration of 20 m/s^2. If after 5secs its engine is switched off , the maximum height of the rocket from the earths surface will be
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Answer:
250m
Explanation:
Given,
Acceleration(a)=20m/s^2
Time(t) =5s
Intial velocity(u) =0
We know,
Distance(s) =ut+1/2at^2
=0+1/2*20*5*5
=250m
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