Question

14. An object thrown vertically up from the ground
passes 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

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Answer 1

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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