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 (see the drawing). 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 0.41 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

v_o = initial velocity of the ball at the time of launch = 15 m/s

= angle of launch of the ball = 50 deg

Consider the motion of the ball along the vertical direction or Y-direction :

Assuming down direction as negative and upward direction as positive

v_oy= initial velocity along the vertical direction = v_o Sin = (15) Sin50 = 11.5 m/s

a_y = acceleration in vertical direction = - 9.8 m/s²

Y = vertical displacement from the point of launch to the point where opponent hits = 2.10 m

t = time taken by the ball = ?

Using the kinematics equation

Y = v_oy t + (0.5) a_y t²

2.10 = 11.5 t + (0.5) (- 9.8) t²

t = 2.15 sec

Consider the motion of the ball along the X-direction or horizontal direction :

v_ox = initial velocity along the horizontal direction = v_o Cos = (15) Cos50 = 9.6 m/s

a_x = acceleration in horizontal direction = 0 m/s²

X = horizontal displacement = ?

t = time of travel = 2.15 s

Using the kinematics equation

X = v_ox t + (0.5) a_x t²

X = (9.6) (2.15) + (0.5) (0) (2.15)²

X = 20.64 m

D = initial distance of the opponent from the ball at the time of launch = 10 m

d = distance to be traveled by the opponent

d = X - D

d = 20.64 - 10

d = 10.64 m

t' = time taken by the opponent to reach the ball and hit it = t - 0.41 = 2.15 - 0.41 = 1.74 sec

minimum average speed of the opponent is given as

v_avg = d/t'

v_avg = 10.64/1.74

v_avg = 6.11 m/s