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 0.38 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.)

To solve this problem, we can break it down into two parts:

Part 1: Find the time it takes for the ball to reach a height of 2.10 m.

To find the time it takes for the ball to reach a certain height, we can use the vertical motion equation:

y = y0 + v0y * t - (1/2) * g * t^2,

where y is the vertical position, y0 is the initial vertical position, v0y is the vertical component of the initial velocity, t is the time, and g is the acceleration due to gravity.

In this case, y = 2.10 m, y0 = 0 m (since the launch point is the reference point), v0y = v0 * sin(theta), where v0 is the initial speed and theta is the launch angle. And g = 9.8 m/s^2.

Plugging in the known values, we get:

2.10 m = 0 m + (15.0 m/s * sin(50.0°)) * t - (1/2) * (9.8 m/s^2) * t^2.

Simplifying the equation, we get:

(1/2) * (9.8 m/s^2) * t^2 - (15.0 m/s * sin(50.0°)) * t + 2.10 m = 0.

Now, solve this quadratic equation for t using the quadratic formula:

t = (-b ± sqrt(b^2 - 4ac)) / (2a),

where a = (1/2) * (9.8 m/s^2), b = -(15.0 m/s * sin(50.0°)), and c = 2.10 m.

Calculating the values, we find that t ≈ 0.579 s or t ≈ 0.091 s (rejecting the negative value).

Therefore, it takes approximately 0.579 seconds for the ball to reach a height of 2.10 m.

Part 2: Find the minimum average speed the opponent must have to reach the ball at that height.

Since the opponent starts moving away from you after a delay of 0.38 s, we need to find the distance the opponent must travel during that time.

Using the equation of motion:

x = x0 + v0x * t,

where x is the horizontal position, x0 is the initial horizontal position, v0x is the horizontal component of the initial velocity, and t is the time.

In this case, x = 10.0 m (distance between you and opponent), x0 = 0 m, v0x = v0 * cos(theta), where v0 is the initial speed and theta is the launch angle. And t = 0.38 s.

Plugging in the known values, we get:

10.0 m = 0 m + (15.0 m/s * cos(50.0°)) * 0.38 s.

Simplifying the equation, we find:

(15.0 m/s * cos(50.0°)) * 0.38 s = 10.0 m.

Now, solve for the horizontal component of the initial velocity:

v0x = 10.0 m / (15.0 m/s * cos(50.0°) * 0.38 s).

Calculating the value, we find that v0x ≈ 3.12 m/s.

Finally, to find the minimum average speed the opponent must have, we need to calculate the speed required to cover the horizontal distance (10.0 m) during the remaining time (0.579 s - 0.38 s = 0.199 s).

Average speed = total distance / total time.

Total distance = horizontal distance (10.0 m).

Total time = time it takes for the ball to reach a height of 2.10 m (0.579 s) - delay time (0.38 s) = 0.199 s.

Therefore, the minimum average speed the opponent must have is:

Average speed = 10.0 m / 0.199 s.

Calculating the value, we find that the opponent must move with a minimum average speed of approximately 50.25 m/s.

To find the minimum average speed that the opponent must move, we will use the concept of projectile motion.

First, let's determine 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:

y = yo + voy * t - (1/2) * g * t^2

where
y = vertical displacement (2.10 m)
yo = initial vertical position (0 m)
voy = initial vertical velocity (vertical component of the initial velocity)
g = acceleration due to gravity (-9.8 m/s^2)
t = time

Since the ball is launched with an initial angle of 50°, the vertical component of the initial velocity is given by:

voy = v * sin(theta)

where
v = initial speed (15.0 m/s)
theta = launch angle (50.0°)

Plugging in the values, we can solve for t:

2.10 = 0 + (15.0 * sin(50.0°)) * t - (1/2) * 9.8 * t^2

2.10 = 7.12t - 4.9t^2

Rearranging the equation, we get:

4.9t^2 - 7.12t + 2.10 = 0

Solving this quadratic equation, we find two possible values for t: t1 = 0.276 s and t2 = 0.773 s. Since the opponent starts moving 0.38 s later, the relevant time for the opponent is t1 + 0.38 s = 0.276 s + 0.38 s = 0.656 s.

Now, let's determine the distance the opponent needs to cover in that time. The horizontal distance covered by the ball in time t is given by:

x = vox * t

where
x = horizontal distance covered (10.0 m)
vox = initial horizontal velocity (horizontal component of the initial velocity)

Since the ball is launched with an initial angle of 50°, the horizontal component of the initial velocity is given by:

vox = v * cos(theta)

Plugging in the values, we can solve for vox:

10.0 = (15.0 * cos(50.0°)) * t

Simplifying the equation, we get:

10.0 = 9.64t

Solving for t, we find t = 1.04 s.

Therefore, the opponent needs to cover the distance of 10.0 m in 0.656 s + 1.04 s = 1.696 s. The average speed is given by:

Average speed = total distance / total time

Average speed = 10.0 m / 1.696 s

Average speed ≈ 5.89 m/s

So, the opponent must move with a minimum average speed of approximately 5.89 m/s.

Note: The above calculation assumes ideal projectile motion without air resistance.