Question

When a ball is thrown vertically upwards it goes trough a dist of 19.6 metres. Find the initial velocity of the ball and time taken by it to rise to the highest point

Answer 1

At maximum height, the velocity of the ball is equal to zero. This means that you an say
v
2
top
 
=
0

=
v
2
0
−
2
⋅
g
⋅
h

, where
h
- the height travelled upward from the initial position.
You get that
h
=
v
2
0
2
⋅
g
=
19.6
2
m
2
s
2
2
⋅
9.8
m
s
2
=
19.6 m
The time it took to reach maximum height, i. e. travel 
h
upward, is
v
top
 
=
0

=
v
0
−
g
⋅
t
up
t
up
=
v
0
g
=
19.6
m
s
9.8
m
s
2
=
2 s
So, it climbs for two seconds, and its total flight time is six seconds, then that must mean that it fell from maximum height in four seconds
t
down
=
t
total
−
t
up
t
down
=
6 s
−
2 s
=
4 s
This means that maximum height, 
H
, is
H
=
v
top
 
=
0

⋅
t
down
+
1
2
⋅
g
⋅
t
2
down
H
=
1
2
⋅
9.8
m
s
2
⋅
4
2
s
2
=
78.4 m
The height of the tower was
h
tower
=
H
−
h
h
tower
=
78.4 m
−
19.6 m
=
58.8 m

Answer 2

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