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 0.38 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.)

Question

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 0.38 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.)

Answer 1

To find the minimum average speed required for your opponent to reach the ball, we need to consider the vertical and horizontal components separately.

Let's start with the horizontal component:

The initial horizontal speed of the ball, also known as the x-component, is given by:
Vx = V_initial * cos(angle)
Vx = 15.0 m/s * cos(50.0°)
Vx = 15.0 m/s * 0.6428
Vx ≈ 9.64 m/s

Now, let's calculate the time it takes for the ball to reach its maximum height:
Using the vertical component, we can find the initial vertical speed of the ball, also known as the y-component:
Vy = V_initial * sin(angle)
Vy = 15.0 m/s * sin(50.0°)
Vy = 15.0 m/s * 0.7660
Vy ≈ 11.49 m/s

The time taken to reach maximum height can be determined by the following formula:
Vy = V_initial_y - g * t_max
11.49 m/s = 15.0 m/s * sin(50.0°) - 9.8 m/s^2 * t_max

Solving for t_max, we get:
t_max = (15.0 m/s * sin(50.0°)) / 9.8 m/s^2
t_max ≈ 1.071 s

Now, let's find the time it takes for your opponent to reach the ball:
Since your opponent starts moving away from the ball 0.38 s later, the total time he has to reach the ball is reduced by that amount:
t_total = t_max - 0.38 s
t_total = 1.071 s - 0.38 s
t_total ≈ 0.691 s

Now, let's calculate the horizontal distance covered by the ball in that time:
The horizontal distance covered by the ball is given by:
x = Vx * t_total
x = 9.64 m/s * 0.691 s
x ≈ 6.67 m

Now we need to determine how far your opponent needs to move to reach the ball. Since he starts 10.0 m away from the ball, the remaining distance to cover is:
d_remaining = x - opponent_initial_distance
d_remaining = 6.67 m - 10.0 m
d_remaining = -3.33 m

Since the remaining distance is negative, it means your opponent overtook the ball before reaching it. Therefore, it is not possible for your opponent to reach the ball and hit it back when it is 2.10 m above its launch point.