# 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 sabantikarty

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by sabantikarty

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so object reaches highest peak in 5s (as it will take equal time interval for going up and coming down)

let velocity at highest point be V

applying equation of motion

V=U+at

here V is final velocity

U is initial velocity

a is acceleration due to gravity(here it is negative acceleration as acceleration due to gravity acts in opposite direction, so a=-g=(-9.8m/s²)

here V=0, a=-9.8m/s², t=5s

0=U-9.8*5

U=49m/s

at 5m height V=49m/s, a=-9.8m/s²,s=5m

V²=U² + 2as

49²=U² - 2*9.8*5

U²=2499

U=49.9899 m/s

Again V=U+a*t

so final velocity V=0, initial velocity U=49.9899m/s, a=-9.8m/s², t we need to find

0=49.9899-9.8*t

t=49.9899/9.8

=5.101s (it is the time of going up)

So total time period is 2*5.101=10.202s