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an object thrown vertically up from groundpasses the height 10 metre twice in the time interval of 10 sec the time of flight is

Singhaniya , 4 Years ago
Grade 12
anser 2 Answers
Sandeep

Last Activity: 4 Years ago

- 2*10*5u= 51 m/stime taken to reach 5m =.1 secso total time of flight = 10.22= u2particle passes height 5mtwice in an interval of 10 sec .So it took 5 sec for reaching heighest point of flight.sovelocity at 5 m height is0= v- gt0=v-10*5v=50so speed at start = 50

Vikas TU

Last Activity: 4 Years ago

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
 

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