ies accordingly. In our case, in order to solve the object tasks it is sufficient, due
to the only momentary nature of this action to consider not the curve shape in the
diagram but only the value of the
where S is the force impulse for the time t a
9.3 Calculating the travelling speeds of the two colliding solid bodies: racket
and bal
U2 = U1 (m-kM) /(M+m
V2 = U1 (1+k)m/(M+m) + (M
K =
K = when a body falls down fr
where: U2 is the ball speed after the strike,
V1- racket speed before strike;
V -racket speed after strike;
h - bouncing height
Or, from (11) above:
V2= 0.060(1+0.35)58.3/0.320+0.06 + (032-0.35.0.06).0/0.32+0.06
.43 m/sec for K=0.35; m=0.06; M=0.32
This means that after the collision the racket will travel in the
at a speed of 12.43 m/sec provided that the p
layer has only received the ball
ut blowing at it [17].
ase after meeting the static (and not clamped
) racket the speed of the
ball in the backward direction will be:
U2 = (0.060-0.35.0.32)58.3/0.32+0.06 + 0.(1.35).0.32/0.32+0.06 =
U2 = -7.98 m/sec
If the player has counter blown at a speed of
V1 = 10 m/sec, then U2= +19.35 m/sec
If we take into account the effort in the handle and the approximate mass of the
