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.)

Vo = 16m/s @ 47 Deg.

Xo = 16*cos47 = 10.91 m/s.
Yo = 16*sin47 = 11.70 m/s.

Tr = (Yf-Yo)/g,
Tr = (0-11.70) / -9.8 = 1.19 s. = Time
to rise to max ht.

hmax = (Yf^2-Yo^2)/2g.
hmax = (0-(11.70)^2) / -19.6 = 6.98 m.
= Max ht above launching point.

d = Vo*t + 0.5g*t^2 = 6.98-2.10,
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 = (23.9-10) m,
0.5a*(2.19-0.30)^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*(2.19-0.30) = 14.7 m/s. =
Speed of plaver.

To find the minimum average speed at which the opponent must move, we need to analyze the motion of the lobbed ball.

Let's break down the problem step-by-step:

Step 1: Determine the initial vertical and horizontal velocities of the lobbed ball.

Given:
Initial speed (Vi) = 16.0 m/s
Launch angle (θ) = 47.0°

Using trigonometry, we can find the initial vertical velocity (Viy) and horizontal velocity (Vix):

Viy = Vi * sinθ
Vix = Vi * cosθ

Viy = 16.0 * sin(47.0°)
Viy ≈ 10.95 m/s (rounded to two decimal places)

Vix = 16.0 * cos(47.0°)
Vix ≈ 10.79 m/s (rounded to two decimal places)

Step 2: Calculate the time it takes for the ball to reach its peak height.

The time it takes for the ball to reach its peak height (t_peak) can be found using the equation:

Viy = Vi * sinθ
Viy = g * t_peak / 2

where g is the acceleration due to gravity (9.8 m/s^2). Rearranging the equation:

t_peak = 2 * Viy / g

t_peak = 2 * 10.95 / 9.8
t_peak ≈ 2.23 s (rounded to two decimal places)

Step 3: Calculate the total time of flight for the lobbed ball.

The total time of flight (t_flight) can be calculated as twice the time it takes for the ball to reach its peak height:

t_flight = 2 * t_peak

t_flight = 2 * 2.23 ≈ 4.47 s (rounded to two decimal places)

Step 4: Calculate the horizontal distance covered by the lobbed ball.

The horizontal distance covered by the lobbed ball (d) can be calculated using the equation:

d = Vix * t_flight

d = 10.79 * 4.47
d ≈ 48.25 m (rounded to two decimal places)

Step 5: Calculate the time it takes for the opponent to reach the ball.

The opponent begins moving 0.30 s after the lob is hit. Therefore, the time it takes for the opponent to reach the ball is:

t_opponent = t_flight - 0.30

t_opponent = 4.47 - 0.30
t_opponent ≈ 4.17 s (rounded to two decimal places)

Step 6: Calculate the minimum average speed of the opponent.

The opponent needs to cover a horizontal distance of 10.0 m to reach the ball. Therefore, the minimum average speed (V_opponent) can be calculated as:

V_opponent = d / t_opponent

V_opponent = 10.0 / 4.17
V_opponent ≈ 2.40 m/s (rounded to two decimal places)

So, the opponent must move with a minimum average speed of approximately 2.40 m/s to reach the ball before it reaches a height of 2.10 m.

To solve this problem, we need to analyze the motion of the lobbed ball and the opponent's movement separately.

First, let's find the time it takes for the ball to reach a height of 2.10 m above its launch point. We can use the vertical motion equation:

h = (v₀sinθ)t - (1/2)gt²

where h is the height, v₀ is the initial velocity, θ is the angle above the horizontal, g is the acceleration due to gravity, and t is the time.

Plugging in the values:
h = 2.10 m
v₀ = 16.0 m/s
θ = 47.0°
g = 9.8 m/s² (acceleration due to gravity)

We can rearrange the equation to solve for t:
t = (h + (1/2)gt²) / (v₀sinθ)

Substituting the given values:
t = (2.10 + (1/2)(9.8)(t²)) / (16.0sin(47.0°))

Now, let's analyze the opponent's movement. The opponent starts moving 0.30 s after you lob the ball. Let's denote the distance covered by the opponent as d and the time taken as T.

We can use the formula for uniformly accelerated linear motion to find T:

d = (1/2)(v₀ + v)T,

where v is the final velocity. Since the opponent is aiming to reach the ball at a certain height, v can be calculated using the equation:

v = gT + v₀sinθ.

The minimum average speed of the opponent can be found using the formula:

v_avg = d / T.

Now, let's combine these equations:

Substituting (1) into (2):
d = (1/2)(v₀ + (gT + v₀sinθ))T
d = (1/2)(2v₀T + gT² + v₀sinθT)

Substituting the expression for d into the equation for v_avg:
v_avg = (2v₀T + gT² + v₀sinθT) / T
v_avg = 2v₀ + gT + v₀sinθ

We want to minimize the average speed of the opponent, so we need to differentiate v_avg with respect to T and set it equal to zero to find the minimum:

dv_avg/dT = g - v₀sinθ = 0

Solving for T:
g = v₀sinθ
T = g / (v₀sinθ)

Now we can substitute this value of T back into the equation for v_avg to find the minimum average speed:

v_avg = 2v₀ + gT + v₀sinθ
v_avg = 2v₀ + gv₀sinθ / (v₀sinθ)
v_avg = 2v₀ + g

Substituting the known values:
v_avg = 2(16.0 m/s) + (9.8 m/s²)
v_avg = 32.0 m/s + 9.8 m/s²
v_avg = 41.8 m/s

Therefore, the opponent must move with a minimum average speed of 41.8 m/s to reach the lobbed ball when it is 2.10 m above its launch point.