A ball is thrown at 15.5 m/s at 42° above the horizontal. Someone located 30 m away along the line3 of the path starts to run just as the ball is thrown. How fast, and in which direction, must the catcher run to catch the ball at the level from which it was thrown?
The range of the projectile is
R(x) = 15.5^2 * sin(84°)/9.81 = 24.36 m
The height is
y(t) = 15.5 sin 42° t - 4.9t^2, so it takes 2.12 seconds to fall back to the initial height.
So, the runner has to run toward the thrower at (30-24.36)/2.12 = 2.66 m/s
To find the speed and direction the catcher must run in order to catch the ball at the same level it was thrown, we need to consider the relative motion between the ball and the catcher.
First, let's break down the initial velocity of the ball into its horizontal and vertical components. The initial velocity given is 15.5 m/s at 42° above the horizontal.
Vertical component of initial velocity: V_y = V_i * sin(θ)
V_y = 15.5 m/s * sin(42°)
V_y ≈ 10.50 m/s
Horizontal component of initial velocity: V_x = V_i * cos(θ)
V_x = 15.5 m/s * cos(42°)
V_x ≈ 11.57 m/s
Now, let's determine the time it takes for the ball to reach the person who starts running. The horizontal distance from the thrower to the person is 30 m.
Horizontal distance = V_x * t
30 m = 11.57 m/s * t
t ≈ 2.59 s
Therefore, the ball takes approximately 2.59 seconds to reach the person.
During this time, the person has also been running. Let's denote the speed of the person as V_p and the distance they run as d (which is 30 m).
The relative velocity between the ball and the catcher must be such that the horizontal distances covered by both are equal. This means:
V_catcher * t = d
V_catcher = d / t
V_catcher = 30 m / 2.59 s
V_catcher ≈ 11.58 m/s
So, the catcher must run at a speed of approximately 11.58 m/s in order to catch the ball at the level it was thrown.
Finally, let's determine the direction in which the catcher must run. Since the catcher needs to match the horizontal speed of the ball, they must run in the same direction as the horizontal component of the ball's velocity.
Therefore, the catcher must run in the direction of the horizontal component of the initial velocity, which was 42° above the horizontal.