A ball is thrown with an velocity u from ground . The ball is at a height of 80m at two times .Time interval being 6s ,find u.

Dear student,
given : initial speed=u
height =80 m
time interval=6 s
let the time taken to ascent=t1 s and the time taken to ascent=t2 s
so, applying equations of motion,
0=u-gt1
and u=gt2
and t1+t2=6
=>2u/g =6
Taking g= 10 ms^-2,
=>u=29.4 m/s [Ans]
Hope this helps.
Thanks,

Soumya Ranjan Mohanty.