(^374) A Textbook of Engineering Mechanics
This is a quadratic equation in t.
∴
- 24.9 (24.9) – 4^2 4.9 (– 20)
0.71 s
2 4.9
t
±××
==
×
Ans.
Example 17.24. Two particles A and B are dropped simultaneously from rest two points
both 100 m above the ground. Particle A falls on the ground, while the particle B in its mid path, hits
a fixed plane inclined to the horizontal as shown in the Fig. 17.3.
Fig. 17.3.
As a result of this impact, the direction of its velocity becomes horizontal. Compare the times
of fall of the particles A and B to reach the ground.
Solution. First of all, consider motion of the stone A (i.e. from A to C). In this case, initial
velocity (u 1 ) = 0 (because it is dropped ) and distance (s 1 ) = 100 m
Let t 1 = Time taken by the particle A to reach C.
We know that distance travelled by the stone A (s 1 )
222
11
11
100 0 9.8 4.9
22
=+ut gt =+×t =t
∴ 1
100
4.52 s
4.9
t ==
Now consider motion of the stone B (first from B to D).
In this case, initial velocity (u 2 ) = 0 (because it is dropped ) and distance (s 2 ) = 50 m
Let t 2 = Time taken by the particle B to reach D.
We know that the distance travelled by the stone B (s 2 )
222
22
11
50 0 9.8 4.9
22
=+ut gt =+×t =t
∴ 2
50
3.19 s
4.9
t == ...(ii)
Now consider motion of the stone B (from D or E to F).
Since the direction of the particle B, after impact at D, becomes horizontal, therefore its
velocity in the vertical direction becomes zero. A little consideration will show that the particle B will
take another 3.19 sec to reach from D to F. Therefore total time taken by the particle B to reach F will
be T = 3.19 + 3.19 = 6.38 sec. Therefore ratio of the two times,
1
6.38
1.41
4.52
T
t
== Ans.
Example 17.25. A cage descends in a mine shaft with an acceleration of 0.5 m/s^2. After the
cage has travelled 25 m, a stone is dropped from the top of the shaft. Determine the (a) time taken by
the stone to hit the cage, and (b) distance travelled by the cage before impact.