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An object thrown vertically up from the ground passes the height 5m twice in an interval of 10 sec. What is its time of flight? Will it be right to take 5 m as the highest point in this question? Please explain.

Aabha Misra , 7 Years ago
Grade 12th pass
anser 1 Answers
Sandeep Bhamoo
Take this height of 5m as it`s lowest point in this interval of 10s ,hence it reaches max. height at 5s and at max. height it`s velocity is zero since v=u`+at. 0=u`-10•5. u`=50m/s , it is the velocity of object at a height of 5m and since v² -u²= 2as. v=u`,u=u°-initial vel. at ground s=5m. putting these values (u°)²=3500. u°=10√35m/s. and time of flight = 2u°/a=(20√35)/(20)= √35s. No it is not right to take 5m as heighest point.
Last Activity: 7 Years ago
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