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 0.35 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.)

The height of the ball is

y = x tanθ - g/2(v cosθ)^2 x^2

so, we have

y = 1.19x - .0527x^2

Now, we want x when y=2.10

1.19x - .0527x^2 = 2.10
x = 20.651

Now, since the ball's horizontal speed is 15 cosθ = 9.642 m/s, that means that it takes 2.14 seconds for the ball to reach the intended spot.

So, the opponent only has 1.79 seconds to move the required 10.651 meters, for a speed of 5.95 m/s

whew

To find the minimum average speed at which the opponent must move to reach the ball, we can break down the problem into horizontal and vertical components.

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 equations:

The equation for vertical displacement is given by:
y = yo + vyo * t + (1/2) * a * t^2

Since the initial vertical velocity of the ball is zero (it is being lofted into the air), the equation becomes:
2.10 m = 0 + (1/2) * (-g) * t^2

where g is the acceleration due to gravity (9.8 m/s^2) and t is the time.

Simplifying the equation, we get:
2.10 m = -4.9 m/s^2 * t^2

Rearranging the equation, we have:
t^2 = 2.10 m / -4.9 m/s^2

t^2 = -0.4286 s^2

Since time cannot be negative, we discard the negative sign and take the positive square root:
t ≈ 0.655 s

Now, let's find the horizontal distance the opponent needs to cover in this time to reach the ball. We can use the horizontal motion equation:

The equation for horizontal displacement is given by:
x = xo + vxo * t

where xo is the initial horizontal position (10.0 m), vxo is the initial horizontal velocity (which is equal to the initial speed of the ball multiplied by cos(θ)), and t is the time.

Given that the initial speed of the ball is 15.0 m/s and the launch angle is 50.0°, we can calculate vxo as:
vxo = 15.0 m/s * cos(50.0°)

vxo = 15.0 m/s * 0.6428

vxo ≈ 9.642 m/s

Plugging in the values, we have:
x = 10.0 m + 9.642 m/s * 0.655 s

x ≈ 16.41 m

Therefore, the opponent needs to cover a horizontal distance of approximately 16.41 m to reach the ball.

Finally, we can calculate the minimum average speed the opponent must move by dividing the horizontal distance by the time taken:
Average speed = (16.41 m) / (0.35 s + 0.655 s)

Average speed ≈ 14.32 m/s

Therefore, the opponent must move with a minimum average speed of approximately 14.32 m/s to reach the ball when it is 2.10 m above its launch point.