Question

Two balls x and y are thrown from top of tower one vertically upward and other vertically downward with same speed. If times taken by them to reach the ground are 6 s and 2 s respectively, then the height of the tower and initial speed of each ball are (g = 10 m/s2)

Answer 1

The ball thrown upwards with speed
u
from top of tower when comes down back to same height (top of tower), will have same speed with which it was thrown upwards i.e.
u
. Further it will take 2 sec to reach ground. This is because the second ball thrown downwards with speed
u
, did take 2 sec. But the total journey took 6 secs. This means ball thrown upwards took 4 secs to reach top, and come back down to top of tower. Which means 2 sec to reach the top where speed became 0
0
, and also exactly when ball thrown downwards reached the ground. (2 sec)

For ball thrown upwards -

=−
v
=
u
−
g
t

0=−(2)
0
=
u
−
g
(
2
)

=2
u
=
2
g
m/s — speed with which balls thrown

For ball thrown downwards (with speed 2
2
g
)

=+122
s
=
u
t
+
1
2
g
t
2

=(2)2+12(2)2
H
=
(
2
g
)
2
+
1
2
g
(
2
)
2

=4+2=6
H
=
4
g
+
2
g
=
6
g
m

=60
H
=
60
m

Edit - Earlier I took a sign wrong in second equation and obtained small height for tower. Soon realized the mistake and corrected the sign.

Answer 2

Answer:

60 metres is the height of the tower.