Question

A person throws a ball vertically upwards with an initial velocity 15m/s.calculate,
1)how high it goes.
2)how long the ball is in air before it comes to his hand?

Answer 1

Initial velocity is 15 m/s.
final velocity at the highest point will be 0 m/s
acceleration due to gravity is 9.8 m/s².
initial velocity (u) is 15 m/sec
Now by the equation of motion :-
v²=u²-2gh
⇒15²=2×9.8×h
⇒h=225/19.6
⇒h=11.479591 metres ( the maximum Height reached ).
Now the return time:
to calculate the return time first we have to calculate the final velocity from when it falls down to the earth; It will be calculated by using the Equation of motion; 
v²=u²+2gh
where Height is 11.479591 m
and Initial velocity (u) is 0m/s
so,
v²=0+2×9.8×11.479591
⇒v=√225
⇒v=15 m/s
Now the to find the time taken to reach the ground we will use the first equation of motion
v=u+gt
u=0
v=15
g=9.8
t=v/g
⇒t=(15/9.8)×2     as it is the return so we will multiply it with 2
⇒t=3.0622 seconds

Answer 2

we have these equations of kinematics:  where acceleration is a constant.
     1    v = u + a t
     2    s = (v + u)/2  * t
     3    v² = u² + 2 a s
     4    s = u t + 1/2 a t²
  
1)     
  we apply the 3rd formula:  with v = 0  and  a = -g = -10 m/s²   and u = 15 m/s
         0 = 15² - 2 *10 * s
         height to which it reaches :  s = 11.25 meters

2)      we apply the formula 1
         t = time to reach the top.   v = 0.   a = -g   
               t = 15/10 = 1.5 sec

the time duration for the ball to reach back the ground is same as the time duration to go up to the top.
     total time duration of the ball in the air / in flight = 2 ( 1.5) = 3 seconds