a stone is thrown upwards with an initial speed of 10m/s. calculate the time after which it acquires the same initial speed but travels in opposite direction.

total time of flight is the time after which it attains same speed but in

opposite direction...

T = 2(u/g)^1/2 = 2sec

Approved

let 'u' be initial velocity=10m/s.

we know that v=u-gt1(t1 is time of ascend, v=0,g=10m/s2)

t1=1 second.

during ascend u=0 and let t2 be time of descend,v=10m/s

by plugging the values in equation v=u+gt2

we get t2=1second

therefore the time after which it aquires same speed is t1+t2=2 seconds

plzzzzzzzz approve!!!!!!!!!!!!!!!