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An aeroplane is flying horizontally at a height of 980 m with velocity 100ms-1 drops a food packet . Aperson on the ground is 414m ahead horizontally from the dropping  point. At what velocity should he move so that he can  catch the food packet
Asked by valavanvino1011 | 12 Aug, 2018, 05:58: PM
When the air plane moving with a speed 100 m/s at 980 m height and dropping the food packet,
then it is equivalent for food packet that it is projected with horizontal velocity 100 m/s at 980 m height
above ground. The time taken by food packet to reach ground is calculated using the formula " S = (1/2)gt2 "
becuase initial vertical component of velocity is zero.

Time taken t for the food packet to reach the ground is given by, 980 = (1/2)×9.8×t2 or  t = 10√2 s .
In this time, horizontal distance travelled by the food packet is = 10√2×100 = 1414 m.

It is given that the person is at 414 m from dropping point. Hence he has to run a distance of 1000 m in 10√2 seconds.
Hence his speed is 1000/(10√2 ) = 70.71 m/s

Answered by Thiyagarajan K | 13 Aug, 2018, 01:58: PM
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