Question

3. A ball is thrown vertically upward with a speed v from a
height h metre above the ground. The time taken for the ball
to hit ground is.....?
​

Answer 1

s=vt−
2
1
​
gt
2

Now, at C (maximum height point),
h
‘
=
2g
v
2

​

So, for journey from C to B,
time =
g
2(h+h
‘
)
​

​
=
g
2(h+
2g
v
2

​
)
​

​
=t
2
​

⇒ total time =t
1
​
+t
2
​
=
g
v
​

Answer 2

Let the upward motion as negative.

Let the height achieved = - h

Velocity of the ball = v

Now using second equation of Motion.

S = ut + 1/2 gt²

-h = vt - 1/2 gt²

-h = (2vt - gt²)/2

gt² - 2vt - 2h = 0

This is in the form of Quadratic equations.

D = b² - 4ac

D = (2v)² - 4 . g . (-2h)

D = 4v² + 8gh

Using Quadratic formula to find the value of 't'.

t = -b + √D/2a

t = -(-2v) + √4v² + 8gh/2g

t = 2v/2g [1 + √1+2gh/v²]

t = v/g [ 1 + √1+2gh/v²] Answer.....