CHAPTER I. MECHANroS
149
28. h ~ 57 m; t ~ 3.4 s, Solution. The equations for the lengths of paths AC and AB (Fig. 194) traversed by the body from the start of fall will be
AC =h ~ gt
2
AB ~==!!... =K (t-1)2
and
222
where t is the time of fall from A to c. Solving these equations gives us the required values of t and h. Second method. Let us consider the equations for the paths AB and Be. The equation for AB is gt~
h
"2=2 where t, is the time of motion of the falling body from A to B. For BC gt~
h
""2==vot 2
where Vo =
-:~ 2g ~
+-t
is the velocity of the body at
Band t 2 = f is the time of motion from B to C. Fig. 194 The total time of fall is t = t 1 t2 = t1 1. By solving these equations we can obtain the values of hand t, H +h (H +h)2 29. Vo ~ 2gH; h m ax = 4H
+
+
---ur- 1/-
If
Solution. The path traversed by the first body before it meets the second is gt 2
H=2 and by the second body before it meets the first is gt 2
h=vot-T After a simultaneous solution of these equations
Vo=
~th 1/2iH
Hence , hmax =
When H
= h we have:
v~ 0= (H Vo =
30• t __1/2gH -1f2gii., g
t
h
4
2g
1/2gh;
)2 (h max
> h)
h max = h.
1/2-h
vO::::::Jg •