A ball in tennis is lofted with an initial speed of 15m/s, at and angle of 50 degrees above the horizonal. At this instant, the opponent is 10m away from the ball. He begins moving backwards 0.3s later, hoping to reach the ball and hit it back at the moment it is 2.10m above its launch point. With what minimum speed must he move?

To find the minimum speed at which the opponent must move, we need to calculate the time it takes for the ball to reach a height of 2.10m and the horizontal distance it travels during that time.

Step 1: Find the time it takes for the ball to reach a height of 2.10m.
Using the kinematic equation for vertical motion:
y = y₀ + v₀y*t - (1/2)gt²
where:
y = final height = 2.10m
y₀ = initial height = 0m (launch point)
v₀y = initial vertical velocity = v₀*sin(θ)
v₀ = initial speed = 15m/s
θ = launch angle = 50°
g = acceleration due to gravity = 9.8m/s²

Substituting the known values:
2.10 = 0 + 15*sin(50°)*t - (1/2)(9.8)t²

Simplifying the equation:
4.9t² - 7.0t + 2.10 = 0

Solving this quadratic equation, we find:
t = 0.78s (rounded to two decimal places)

Step 2: Find the horizontal distance the ball travels during that time.
Using the kinematic equation for horizontal motion:
x = x₀ + v₀x*t
where:
x = horizontal distance
x₀ = initial horizontal position = 10m
v₀x = initial horizontal velocity = v₀*cos(θ)

Substituting the known values:
x = 10 + 15*cos(50°)*0.78

Simplifying the equation:
x = 21.40m (rounded to two decimal places)

Step 3: Find the minimum speed the opponent must move.
The opponent needs to cover a horizontal distance of x = 21.40m during the time t = 0.78s to reach the ball.

Minimum speed = horizontal distance / time
Minimum speed = 21.40m / 0.78s

Calculating this, we find:
Minimum speed = 27.44m/s (rounded to two decimal places)

Therefore, the opponent must move at a minimum speed of 27.44m/s to reach the ball and hit it back at the desired height.