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.)
See your 6:20pm post for solution.
To find the minimum average speed at which your opponent must move, we need to determine the time it takes for the ball to reach a height of 2.10 m above its launch point.
First, let's break down the initial velocity into its horizontal and vertical components. The initial speed of 16.0 m/s makes an angle of 47.0° with the horizontal. The horizontal component (Vx) can be found using the formula Vx = V * cos(θ) and the vertical component (Vy) using Vy = V * sin(θ), where V is the initial speed and θ is the launch angle.
Vx = 16.0 m/s * cos(47.0°)
Vx ≈ 10.89 m/s
Vy = 16.0 m/s * sin(47.0°)
Vy ≈ 11.53 m/s
Now, let's find the time (t) it takes for the ball to reach a height of 2.10 m above its launch point. We will use the vertical motion equation: h = Vy * t - 0.5 * g * t^2, where h is the height, t is the time, and g is the acceleration due to gravity (-9.8 m/s^2).
2.10 m = 11.53 m/s * t - 0.5 * 9.8 m/s^2 * t^2
This is a quadratic equation. Rearranging and simplifying, we get:
4.9 * t^2 - 11.53 * t + 2.10 = 0
Now, we can solve this quadratic equation using the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a), where a = 4.9, b = -11.53, and c = 2.10.
t = [-(-11.53) ± √((-11.53)^2 - 4 * 4.9 * 2.10)] / (2 * 4.9)
Simplifying further, we obtain:
t ≈ 1.50 s or t ≈ 0.103 s
Since we are interested in the minimum average speed, we will consider the shorter time of 0.103 s.
Now, let's determine the horizontal distance (d) traveled by your opponent during this time. We can use the equation d = Vx * t, where Vx is the horizontal component and t is the time.
d = 10.89 m/s * 0.103 s
d ≈ 1.12 m
Therefore, your opponent must move at least 1.12 meters in 0.30 seconds to reach the ball and hit it back when it is 2.10 m above its launch point.
To find the minimum average speed, we divide the distance traveled by the time taken:
v = d / t
v = 1.12 m / 0.30 s
v ≈ 3.73 m/s
Therefore, your opponent must move with a minimum average speed of approximately 3.73 m/s.