If a ball is projected vertical upward from ground such that total distance traveled is 3.6 times of distance travel by the ball in `1​^(st)`​ second, then find time of flight of ball ?

Options

A. `2.4s`
B. `3s`
C. `6s`
D. `4.8s`

Answer

C

Explanation

`d_("total") = 2h = 3.6s_(1),h=1.8 s_(1)`

t= time of ascent
t= times of descent
`h=(1)/(2) gt^(2)` …(i)
`h- s_(1)=(1)/(2)g(t-1)^(2)…(2)`
by (1) & (2) `(1.8s_(1))/(1.8s_(1))-s_(1)=(t^(2))/((t-1)^(2))`
`implies (t^(2))/((t-1)^(2))=(1.8s_(1))/(0.8s_(1))=9/4`
`implies (t)/(t-1)=(3)/(2)implies t=3`
time of flight = `2t=6 sec `