Question

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?

Answer 1

object passes through 5m height twice in 10s

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