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An object thrown vertically up from the ground passes the height **5 m** twice in an interval of **10 s. **What is the **TIME OF FLIGHT?**

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by fblpker

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by fblpker

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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

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