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An aeroplane takes off at an angle or `30^(@)` to the horizontal. If the component of its velocity along the horizontal is `240 km h^(-1)`. What is the actual velocity. ? Also find the vertical component of its velocity ? |
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Answer» Let (v) be the actual velocity of aroplane while taking off. As per question ` u cos 30^(@) =240` or 1 u= (240)/( cos 30^(@) =9240)/(aqrt 3//2) =(480)/(sqrt 3) = (480 sqrt)/3` `= 160 sqrt 3 km h^(-1)` Vertical component velocity of aeroplace ` =u sin 30^(@) = 160 sqrt 3 xx 1//2` `=80 sqrt 3 km h^(-1)`. |
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