Question

An object thrown vertically up from the ground
passes the height 10 m twice in the time interval
of 10 s. The time of flight is (g = 10 m/s2)
(1) 6/3 s
(2) 313 5
(3) 2/3s
(4) 13 S​

Answer 1

Answer:

The correct answer will be 9.59 sec

Explanation:

let v' be the velocity of the particle

the new time of flight is 10 sec

let t is the time of flight

t= 2v'/g

=> v'= gt/2

=> v'= 10 x 10 /2

=> v' = 50 m/s

Now the distance between the object and ground is 10 m

Let the actual velocity of projection is v

v'^2= v^2-2gh

=> 50^2= v^2- 2 x 10 x 10

=> v^2= 2300

=> v= √2300

       = 47.95 m/s

hence the actual time of flight, t= 2 x 47.95/10= 9.59 sec