Question

a ball is thrown vertically upward with a velocity of 19.6 m/s find maximum height reached by the ball and time taken to reach maximum height

Answer 1

dear friend
if ball reach maximum height
then velocity =0
use v=u+at
0=19.6-gt
t=19.6/9.8=2sec
now again use v^2=u^2+2as
where v=0
u=19.6m/sec
a= -9.8m/s^2
h is the maximum height
if you solve
find h =u^2/2g=(19.6)^2/(2 x 9.8)=19.6m

Answer 2

See the image:

Given,
u = 19.5 m/s
t = 5 sec
g = 9.8m/s²
Total distance covered while going up = S
Total distance covered while coming down = h+S
At topmost points
v= 0
v= u - gt
0 = 19.5-10(9.8)
t= 2 sec (approx.)
Now,
the velocity of ball when it falls down a distance S after reaching topmost height (h+S) will be same as initial velocity and the time taken will be again 2 sec.
So, total time elapsed = time for going up S + time for coming down S = 2 +2 = 4
Total time = 5 sec.
Renaming time = 5-4 = 1 sec.
here u = 19.6m/s(downwards)
t= 1 sec
h = ut +1/2 gt²
h = 19.5 (1) + 1/2 (9.8) (1) = 19.5 + 4.9 = 24.4m