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An aeroplane takes off at an angle of 45 0 to the horizontal. If the vertical component of its velocity is 300 kmph, calculate its actual velocity. What is

the horizontal component of velocity?



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Vertical component of any projectile type motion is u sinθ
so given u sin 45° = 300
so u = 300√2 km/hr
  u = 424.2 km/hr 
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//As we know 45° to horizontal is same as 45° to vertical// vertical angel 45° is given speed=V sin45 300= V sin45 V= 300 X 1.4131 = 423.93 kmph ~ 424 kmph
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