(^370) „„„„„ A Textbook of Engineering Mechanics
Solution. First of all, consider the upwards motion of the first stone. In this case, initial
velocity (u) = – 49 m/s (Minus sign due to upward motion) and final velocity (v) = 0 (because stone
is at maximum height)
Let t= Time taken by the stone to reach maximum height.
We know that final velocity of the stone (v),
0= u + gt = – 49 + 9.8 t ...(Minus sign due to upwards motion)
∴ t=
49
9.8
= 5 s
It means that the stone will take 5 s to reach the maximum height and another 5 s to come
back to the ground.
∴ Total time of flight = 5 + 5 = 10 s
Now consider the motion of second stone. We know that time taken by the second stone for
going upwards and coming back to the earth
= 10 – 2 = 8 s
and time taken by the second stone to reach maximum height
8
4s
2

Now consider the upward motion of the second stone. We know that final velocity of the
stone (v),
0=u + gt = – u + 9.8 × 4 = – u + 39.2
∴ u= 39.2 m/s Ans.
Example 17.18. A stone is dropped from the top of a tower 50 m high. At the same time,
another stone is thrown upwards from the foot of the tower with a velocity of 25 m/s. When and where
the two stones cross each other?
Solution. Given : Height of the tower = 50 m
Time taken by the stone to cross each other
First of all, consider downward motion of the first stone. In this case, initial velocity (u) = 0
(beacuse it is dropped)
Let t = Time taken for the stones to cross each other.
We know that distance traversed by the stone,
(^11222) 00.5
22
sut=+gt=+gt=gt ...(i)
Now consider upward motion of the second stone. In this case, initial velocity = – 25 m/s
(Minus sign due to upward) and distance traversed = 50 – s,
We know that the distance traversed,
50 –^122 – 25 0.5
2
sut=+gt= t+gt ...(ii)
Adding equations (i) and (ii),
50 = 25 t or^50 2s
25
t== Ans.
Point where the stones crosss each other
Substituting the value of t = 2 in equation (i),
s = 0.5 × 9.8 (2)^2 = 19.6 m Ans.