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14. An object thrown vertically up from the groundpasses the height 5 m twice in an interval of 10 s.What is its time of flight?(1) √28 s(2) √86 s(3) √104 s(4)√72s​

Answer»

Answer:

Let the height of the highest point that the object reached be S meters

The object has climbed the height of (S-5) m in 10s/2 = 5 seconds and it DESCENDED (S-5) m in the same TIME = 5sec

we USE s = ut + 1/2 a t²

S - 5 = 0 * 5 + 1/2 * 9.8 * 5²

S = 127.5 meters

Time duration for the object to travel VERTICALLY 127.5 meters

127.5 = 0 t + 1/2 * 9.8 * t²

t² = 26.02 t = 5.1 sec

Total time of flight = 2 * 5.1 = 10.2 seconds

Explanation:

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