Question

a plane covers 500m in direction of the 60° with runway while starting the flight . find the distance covered by plane in horizontal and vertical direction ​

Answer 1

Answer:

From h=21gt2

we have tOB=g2hOA=9.82×1960=20s

Horizontal distance AB=vtOB

=(600×185m/s)(20s)

answer =3333.33m=3.33km

Explanation:

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Answer 2

$\huge \underline \green{\sf{ Answer :- }}$

• From h = 1/2gt^2

we have tOB = √(2hOA/g)

= √(2 × 1960/9.8)

= 20 s

• Horizontal distance AB = vtOB

= ( 600 × 5/18m/s ) (20s)

= 3333.33m

= 3.33 km