Posted by Natasha on Wednesday, February 8, 2012 at 6:20pm.
The lob in tennis is an effective tactic when your opponent is near the net. It consists of lofting the ball over his head, forcing him to move quickly away from the net (see the drawing). Suppose that you lob the ball with an initial speed of 16.0 m/s, at an angle of 47.0° above the horizontal. At this instant your opponent is 10.0 m away from the ball. He begins moving away from you 0.30 s later, hoping to reach the ball and hit it back at the moment that it is 2.10 m above its launch point. With what minimum average speed must he move? (Ignore the fact that he can stretch, so that his racket can reach the ball before he does.)

physics  Henry, Friday, February 10, 2012 at 10:57am
Vo = 16m/s @ 47 Deg.
Xo = 16*cos47 = 10.91 m/s.
Yo = 16*sin47 = 11.70 m/s.
Tr = (YfYo)/g,
Tr = (011.70) / 9.8 = 1.19 s. = Time
to rise to max ht.
hmax = (Yf^2Yo^2)/2g.
hmax = (0(11.70)^2) / 19.6 = 6.98 m.
= Max ht above launching point.
d = Vo*t + 0.5g*t^2 = 6.982.10,
0 + 4.9t^2 = 4.88,
t^2 = 0.996,
Tf = 1.0 s. = Time to fall to 2.10 m
above launching point.
Dx = Xo(Tr+Tf) = 10.91(2.19) = 23.9 m.
= Hor. dist. of ball.
d = 0.5a*t^2 = (23.910) m,
0.5a*(2.190.30)^2 = 13.9,
0.5a*3.57 = 13.9,
1.786a = 13.9,
a = 13.9 / 1.786 = 7.78 m/s^2. = acceleration of player.
V = at = 7.78*(2.190.30) = 14.7 m/s. =
Speed of plaver.
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