Question

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 . Suppose that you loft the ball with an initial speed of 15.0 m/s at an angle of 50.0° the horizontal. At this instant your opponent is 10.0 m away from the ball. He begins moving away from you 0.21 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.)

Answer 1

Initial speed is 15.0 m/s at 50^o above horizontal.

The vertical component of the initial velocity (Vo)

Vo=15sin(50^o)=11.5 m/s

The horizontal component = 15cos(50^o)=9.6 m/s

the ball takes on this journey is the time it takes to reach its peak and fall to the ground until it is 2.10 m above the ground.

Time it takes to peak is found from the following;
V= Vo-gt
At peak V=0
0=11.5-9.81xt
t=1.17 m/s

The height is found from this equation;
h= h_o+V_ot -1/2(gt²)

h=0+11.5 x1.17- 0.5x(9.81)(1.17)²=13.46-6.71

h=6.75m

To make the ball fall the distance of 6.75 to 2.10m (4.65 m)
distance, d= 0.5xgxt²
4,65=0.5x(9.81)x t²

t=0.97sec

t= 0.97 + 1.17 =2.14 sec.

The Distance the ball goes is from the following eqn;
distance, d= Vxt
d=9.65x2.14=20.65 m
This player has a reaction time of 0.21 s before he starts to go back.

The time is 2.14 s-0.21 s=1.78 s

The speed the guy must go is change in distance/ change in time

speed=20.6m/1.78 s=11.5 m/s