14
1 Kinematics and Statics
1.2 When the stone reaches a height h above A
v
2
1 = u
2
− 2gh
(1)
and when it reaches a distance h below A
v
2
2 = u
2
+ 2gh
(2)
since the velocity of the stone while crossing A on its return journey is again u
vertically down.
Also, v 2 = 2v 1 (by problem)
(3)
Combining (1), (2) and (3) u 2 =
10
3 gh
(4)
Maximum height
H =
u 2
2g
=
10
3
gh
2g
=
5h
3
1.3 Let the stones meet at a height s m from the earth after t s. Distance covered by
the first stone
h − s =
1
2
gt
2
(1)
where h = 19.6 m. For the second stone
s = ut =
1
2
gt
2
(2)
v
2
= 0 = u
2
− 2gh
u =
2gh =
√
2 × 9.8 × 19.6 = 19.6 m/s
( 3 )
Adding (1) and (2)
h = ut, t =
h
u
=
19.6
19.6
= 1 s
1 Kinematics and Statics
1.2 When the stone reaches a height h above A
v
2
1 = u
2
− 2gh
(1)
and when it reaches a distance h below A
v
2
2 = u
2
+ 2gh
(2)
since the velocity of the stone while crossing A on its return journey is again u
vertically down.
Also, v 2 = 2v 1 (by problem)
(3)
Combining (1), (2) and (3) u 2 =
10
3 gh
(4)
Maximum height
H =
u 2
2g
=
10
3
gh
2g
=
5h
3
1.3 Let the stones meet at a height s m from the earth after t s. Distance covered by
the first stone
h − s =
1
2
gt
2
(1)
where h = 19.6 m. For the second stone
s = ut =
1
2
gt
2
(2)
v
2
= 0 = u
2
− 2gh
u =
2gh =
√
2 × 9.8 × 19.6 = 19.6 m/s
( 3 )
Adding (1) and (2)
h = ut, t =
h
u
=
19.6
19.6
= 1 s