A ball is thrown upwards with a speed u from a height . . .

Question : A ball is thrown upwards with a speed u from a height h above the ground. What is the time taken by the ball to hit the ground?

Doubt by Avani

Solution : 

Method 1 : 
Here 

Initial Velocity = u
Total Displacement covered (s) = -h
Let the total time taken (t) = t

Using Second Equation of Motion 
s=ut+½ at²
-h=ut+½ (-g)t²
Multiply both sides by 2
-2h=2ut-gt²
0=2h+2ut-gt²
gt²-2ut-2h=0
This is a quadratic equation
Here
A=g
B=-2u
C=-2h
D=B²-4AC
D=(-2u)²-4(g)(-2h)
D=4u²+8gh

Using Quadratic Formula 

Considering the positive value of t

This is the required time taken by the ball to hit the ground. 

Method 2 : 

Here 
Consider the motion of the body from the point of projection to the highest point.  

Initial Velocity = u (Given)
At the highest point 
Final velocity (v) = 0
Let t₁ be the time taken by the body to reach the highest point. 
Using the First Equation of Motion 
v=u+at
0=u+(-g)t₁
-u=-gt₁
u/g = t₁
t₁ = u/g — (1)

Displacement covered by the body in time t1
v²-u²=2as
(0)²-u²=2(-g)s'
-u²=-2gs'
u²/2g=s'
s'=u²/2g 

Now consider the motion of the body from the highest point the ground. 

Let t2 be the time taken by the body in going from the highest point to the ground. 

Here 
s=s'+h
s=u2/2g+h

a=g

u=0

Using Second Equation of Motion

Hence, the total time taken by the ball to hit the ground