A ball thrown up verically returns to the thrower after 6 s. Find
(a) the velocity with which it was thrown up. (b) the maximum height it reaches, and (c ) its position after 4 s.

Explanation

Given : Total time = 6s
Time of ascent = time of descent
`=6/2 = 3s`
To find : (i) Initial velocity (u)
(ii) Maximum height (`h_1`)
Formula : (i) v=ut+at
`(ii) s=ut+1/2at^2`
Calculation :
(i) For upward motion of the ball ,
v=0, t=3, a=`-9.8m//s^2`
From formula (i) ,
`therefore v=u+at`
`therefore 0=u-9.8xx3`
`therefore u=29.4 m//s`
(ii) For ball moving upwards and reaching maximum height ,
`u=29.4m//s , a=-9.8m//s^2, t=3s, s=h_1`
From formula (ii),
`h_1 = 29.4xx3 -1/2xx9.8xx3^2`
`=88.2 - 44.1`
`therefore h_1 = 44.1m`