Question

when a ball is thrown vertically upwards, it goes through a distance of 19.6 m . find the initial velocity of the ball and the time taken by it to rise to the highest point . the value of g is 9.8 M per second square

Answer 1

the ball goes to distance of 19.6
hence highest point is at distance 19.6/2
=9.8
Now
velocity at highest point is 0
hence
and
v=u+at
0=u-9.8t
9.8t=u. .......1
now
s=ut +1/2at^2
9.8=(9.8t)t+4.9t^2
$1 = {t}^{2} - \frac{1}{2} {t}^{2}$

t=2√2sec
u=9.8t
=9.8√2 m/s

Answer 2

initial velocity-?
final velocity -0
height -19.6m
g -9.8m^2
time -?
{v^2=u^2 +2as}
Put the values in this equation.
0=u^2+2(-9.8)(19.6) -------- we have taken negative value of acceleration.
0=u^2 -(19.6) (19.6)
u^2=(19.6)^2
u =19.6 m

{v=u+at}
So,
0=19.6+(-9.8)(t)
0=19.6-9.8t
time =19.6/9.8
time=2 seconds

Hope it's helpful