08. A jet plane starts from rest with an acceleration of 3ms-2 and makes a run for 35s before taking off. What is the minimum length of the runway and what is the velocity of the jet at take off?.
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Answer: Given: u= 0m/s
a=3m/s-2
t=35s
s=?
v=?
Explanation: s=ut+1/2*a*t2
=0*35+1/2*3*1225
=0+3*612.5
=1837.5
Therefore, distance or the minimum length of runway is 1837.5 meters
Now, v=u+at
=0+3*35
=105m/s
Therefore, the velocity of the jet to take off is 105m/s.
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