Solutions 111
segments are equidistant on the time axis, their
separations being ti = 21/2/ 1 /g for the first ball
and t^2 = 2172h^2 /g for the second ball. Since
hi = 4/i^2 by hypothesis, ti = 2t^2 , i.e. the fre-
quency of motion of the second ball is twice as
high as that of the first ball. It follows from the
ratio of the initial heights that the maximum ve-
locities attained by the balls will also differ by a
factor of two (see Figs. 122 and 123):
vimax = 2v2max = 172hig= vo.
There are two possibilities for the coincidence
of the velocities of the balls in magnitude and di-
rection. The velocities of the balls may coincide
for the first time x = nt 1 s after the beginning of
motion (where n = 0, 1, 2,.. .) during the time
interval t 1 /4, then they coincide 3t 1 /4 s after the
beginning of motion during the time interval ti/2.
Subsequently, the velocities will coincide with a
period t 1 during the time interval t 1 /2. The other
possibility consists in that the second ball starts
moving r = t 1 /2 nt 1 s (where n = 0, 1, 2,.. .)
after the first ball. After t 1 /4 s, the velocities of
the balls coincide for the first time and remain iden-
tical during the time interval t 1 /2. Subsequently,
the situation is repeated with a period t 1.
For other starting instants for the second ball,
the velocity graphs will have no common points
upon superposition because of the multiplicity of
the periods of motion of the balls, and the problem
will have no solution.
1.7*. Let us consider the motion of a ball falling
freely from a height H near the symmetry axis start-
ing from the moment it strikes the surface. At the
moment of impact, the ball has the initial velocity
vo = j/2gH (since the impact is perfectly elastic),
and the direction of the velocity vo forms an angle
2a with the vertical (Fig. 124).
Let the displacement of the ball along the hori-
zontal in time t after the impact be s. Then
co sin 2a•t = s. Hence we obtain t = s/(j/2gH X
sin 2a), where vo sin 2a is the horizontal corn-