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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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
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210
hope this helps you.....
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