Question

A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 15m/s when the hand is 2.0m above the ground. How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.)

Answer 1

Upward:
v = v o - g t
0 m/s = 15 m/s - 9.81 m/s² · t
15 m/s = 9.81 m/s² · t
t = 15 m/s : 9.81 m/s²
t 1 = 1.53 s
h = v o · t - g t² / 2  + 2 m
h = 22.95 m - (9.81 m/s² · 2.3409 s²) / 2 + 2 m
h = 11.475 m + 2 m = 13.475 m
Downward:
13.475 m = 9.81 · t² / 2
t = √(13.475 · 2 ) / 9.81
t 2 = 1.66 s
t = t 1 + t 2 = 1.53 s + 1.66 s = 3.19 s 
Answer: The ball is 3.16 seconds in the air before it hits the ground.

Answer 2

 The ball's initial speed = Vi = 15 m/s 
 Vf = ? 
Acceleration due to gravity = g = 9.8 m/s^2 
Distance ball upward = S.u = ? 
Distanc ball downward = S.d = ? 
Time taken upward = t.u 
Time taken downward = t.d 
Total Time to reach ground = T =?
 upward distance
2gS = Vi^2 
2*(9.8)*S = ( 15 )^2 
19.6 * S = 225 
S = 225 / 19.6 
S = 11.48 meters 
 downward distance..
Upward distance + 2 m = Downward distance
=11.48 m + 2 m 
=13.48 m
 find upward time for the bal
Vi / g = t.u 
15 / 9.8 = t.u 
1.53 seconds = t.u 
downward time
S.d = 1/2*g*t.d ^2 
13.48 = 1/2*9.8 * t.d^2 
13.48 = 4.9 * t.d^2 
13.48 / 4.9 = t.d^2 
2.751 seconds = t.d^2 
Taking square root both sides 
sqrt ( 2.751 ) = sqrt ( t.d^2 ) 
1.658 seconds = t.d 
Total time taken by ball = upward time + downward time 
T = t.u + t.d 
T = 1.53 s + 1.658 s 
T = 3.188 seconds
hope this helps